Artigos

2.E: Equações de Primeira Ordem (Exercícios)


Estes são exercícios de lição de casa para acompanhar o mapa de texto "Equações diferenciais parciais" de Miersemann. Este é um livro didático voltado para um primeiro curso de um semestre sobre equações diferenciais, voltado para estudantes de engenharia. Equações diferenciais parciais são equações diferenciais que contêm funções multivariáveis ​​desconhecidas e suas derivadas parciais. O pré-requisito para o curso é a sequência de cálculo básico.

Q2.1

Suponha que (u: mathbb {R} ^ 2 mapsto mathbb {R} ^ 1 ) seja uma solução de
$$
a (x, y) u_x + b (x, y) u_y = 0.
$$
Mostre que para (H in C ^ 1 ) arbitrário também (H (u) ) é uma solução.

Q2.2

Encontre uma solução (u não equiv const. ) De
$$
u_x + u_y = 0
$$
de tal modo que
$$
mbox {gráfico} (u): = {(x, y, z) in mathbb {R} ^ 3: z = u (x, y), (x, y) in mathbb { R} ^ 2 }
$$
contém a linha reta ((0,0,1) + s (1,1,0), s in mathbb {R} ^ 1 ).

Q2.3

Seja ( phi (x, y) ) uma solução de
$$
a_1 (x, y) u_x + a_2 (x, y) u_y = 0 .
$$
Prove que as curvas de nível (S_C: = {(x, y): phi (x, y) = C = const. } ) São curvas características, desde que ( nabla phi not = 0 ) e ((a_1, a_2) não = (0,0) ).

Q2.4

Prove a proposição 2.2.

Q2.5

Encontre duas soluções diferentes para o problema do valor inicial

[u_x + u_y = 1, ]

onde os dados iniciais são (x_0 (s) = s ), (y_0 (s) = s ), (z_0 (s) = s ).

Dica: ((x_0, y_0) ) é uma curva característica.

Q2.6

Resolva o problema do valor inicial
$$
xu_x + yu_y = u
$$
com dados iniciais (x_0 (s) = s, y_0 (s) = 1 ), (z_0 (s) ), onde (z_0 ) é dado.

Q2.7

Resolva o problema do valor inicial
$$
-xu_x + yu_y = xu ^ 2,
$$
(x_0 (s) = s, y_0 (s) = 1 ), (z_0 (s) = mbox {e} ^ {- s} ).

Q2.8

Resolva o problema do valor inicial
$$
uu_x + u_y = 1,
$$
$ x_0 (s) = s, y_0 (s) = s $, (z_0 (s) = s / 2 ) se (0

Q2.9

Resolva o problema do valor inicial
$$
uu_x + uu_y = 2,
$$
(x_0 (s) = s, y_0 (s) = 1 ), (z_0 (s) = 1 + s ) se (0

Q2.10

Resolva o problema do valor inicial (u_x ^ 2 + u_y ^ 2 = 1 + x ) com dados iniciais fornecidos (x_0 (s) = 0, y_0 (s) = s, u_0 (s) = 1,
p_0 (s) = 1, q_0 (s) = 0 ), (- infty

Q2.11

Encontre a solução ( Phi (x, y) ) de
$$
(x-y) u_x + 2yu_y = 3x
$$
de modo que a superfície definida por (z = Phi (x, y) ) contém a curva
$$
C: x_0 (s) = s, y_0 (s) = 1, z_0 (s) = 0, s in { mathbb R}.
]

Q2.12

Resolva o seguinte problema inicial de cinética química.
$$
u_x + u_y = left (k_0e ^ {- k_1x} + k_2 right) (1-u) ^ 2, x> 0, y> 0
$$
com os dados iniciais (u (x, 0) = 0, u (0, y) = u_0 (y) ), onde (u_0 ), (0

Q2.13

Resolva o problema Riemann
begin {eqnarray *}
u_ {x_1} + u_ {x_2} & = & 0
u (x_1,0) & = & g (x_1)
end {eqnarray *}
em ( Omega_1 = {(x_1, x_2) in mathbb {R} ^ 2: x_1> x_2 } ) e em ( Omega_2 = {(x_1, x_2) in mathbb { R} ^ 2: x_1 Onde
$$
g (x_1) = left { begin {array} {r @ { quad: quad} l}
u_l & x_1 <0
u_r & x_1> 0
end {array} right.
$$
com constantes (u_l not = u_r ).

Q2.14

Determine o ângulo de abertura do cone de Monge, ou seja, o ângulo entre o eixo e o apótema (em alemão: Mantellinie) do cone, para a equação
$$
u_x ^ 2 + u_y ^ 2 = f (x, y, u),
$$
onde (f> 0 ).

T2.15

Resolva o problema do valor inicial
$$
u_x ^ 2 + u_y ^ 2 = 1,
$$
onde (x_0 ( theta) = a cos theta, y_0 ( theta) = a sin theta, z_0 ( theta) = 1,
p_0 ( theta) = cos theta ), (q_0 ( theta) = sin theta ) se (0 le theta <2 pi ),
(a = const.> 0 ).

T2.16

Mostre que a integral ( phi ( alpha, beta; theta, r, t) ), veja o problema de Kepler, é uma integral completa.

Q2.17

a) Mostre que (S = sqrt { alpha} x + sqrt {1- alpha} y + beta ), ( alpha,
beta in mathbb {R} ^ 1, 0 < alpha <1 ), é uma integral completa de (S_x- sqrt {1-S_y ^ 2} = 0 ).
b) Encontre o envelope desta família de soluções.

T2.18

Determine o comprimento do meio eixo da elipse
$$
r = frac {p} {1- varejpsilon ^ 2 sin ( theta- theta_0)}, 0 le varejpsilon <1.
]

T2.19

Encontre a função de Hamilton (H (x, p) ) da equação diferencial de Hamilton-Jacobi-Bellman se (h = 0 ) e (f = Ax + B alpha ), onde
(A, B ) são matrizes constantes e reais, (A: mathbb {R} ^ m mapsto mathbb {R} ^ n ), (B ) é um real ortogonal (n times n ) - Matriz e (p in mathbb {R} ^ n ) são fornecidos. O conjunto de controles admissíveis é dado por
$$
U = { alpha in mathbb {R} ^ n: sum_ {i = 1} ^ n alpha_i ^ 2 le1 } .
]

Observação. A equação de Hamilton-Jacobi-Bellman é formalmente a equação de Hamilton-Jacobi (u_t + H (x, nabla u) = 0 ), onde a função de Hamilton é definida por
$$
H (x, p): = min _ { alpha in U} left (f (x, alpha) cdot p + h (x, alpha) right),
$$
(f (x, alpha) ) e (h (x, alpha) ) são dados. Veja, por exemplo, Evans [5], Capítulo 10.


2.E: Equações de Primeira Ordem (Exercícios)

Expresso [ begin x_1 & # 39 & amp = 2tx_1 + e ^ tx_2 x_2 & # 39 & amp = 3x_1 - 3t x_2 fim] com (x_1 (0) = -5 ) e (x_2 (0) = 2 ) como uma equação vetorial com uma condição inicial do vetor.

Solução

Usando a definição de multiplicação de matrizes, podemos facilmente concluir [ xBld ^ prime = begin 2t & amp e ^ t 3 & amp -3t end xBld ] com condição inicial ( xBld (0) = begin -5 2 fim) .

Exercício 2

Considere o sistema [ begin x & # 39 & amp = 2x + y - z y & # 39 & amp = x-3y + 5z z & # 39 & amp = 4x -7y + z. end] Escreva este sistema como uma equação vetorial.

Solução

Usando a definição de multiplicação de matrizes é facilmente visto que [ frac < dee> < dt> begin x y z end = begin 2 & amp 1 & amp -1 1 & amp -3 & amp 5 4 & amp -7 & amp 1 end começar x y z end]

Exercício 3

Considere a equação vetorial (< bf x> & # 39 = begin 4t & amp 6t ^ 2 2 & amp t ^ 3 end < bf x> + begin e ^ t e ^ <-t> end). Escreva esta equação como um sistema de 2 equações.

Solução

Escrevendo ( xBld = begin x_1 x_2 end). Vemos pela definição de multiplicação de matrizes que [ begin x_1 ^ prime & amp = 4tx_1 + 6t ^ 2x_2 + e ^ t x_2 ^ prime & amp = 2x_1 + t ^ 3x_2 + e ^ <-t>. end]

Para os Problemas 4-7, reformule as equações diferenciais lineares de ordem superior como um sistema linear de equações de primeira ordem. Encontre a matriz de coeficientes ( ABld (t) ) e forçando ( fBld (t) ). Se o problema for um problema de valor inicial, certifique-se de declarar a condição inicial.

Exercício 4

Solução

Defina [ xBld = begin x_1 x_2 end = begin u u ^ prime end] Tomando a derivada de ambos os lados e usando o sistema linear, encontramos [ xBld ^ prime = begin x_2 -3tx_1 + e ^ fim = begin 0 & amp 1 -3t & amp 0 end xBld + begin 0 e ^ t fim.]

Exercício 5

Solução

Defina [ xBld = begin x_1 x_2 x_3 end = begin y y ^ prime y ^ < prime prime> end. ] Tomando a derivada de ambos os lados e usando o sistema linear, encontramos [ xBld ^ prime = begin x_2 x_3 -2x_3 + tx_2 - x_1 end = begin 0 & amp 1 & amp 0 0 & amp 0 & amp 1 -1 & amp t & amp -2 end xBld. ]

Exercício 6

Solução

Exercício 7

Solução

Defina [ xBld = begin x_1 x_2 x_3 x_4 end = begin y y ^ prime y ^ < prime prime> y ^ < prime prime prime> end. ] Tomando a derivada de ambos os lados e usando os EDOs, vemos [ xBld ^ prime = begin x_2 x_3 x_4 -t ^ 2x_4 - cos <(t)> x_3 - t ^ 2 sin <(t)> x_1 + te ^ fim= ABld (t) xBld + fBld (t). ] Onde [ ABld (t) = begin 0 & amp 1 & amp 0 & amp 0 0 & amp 0 & amp 1 & amp 0 0 & amp 0 & amp 0 & amp 1 -t ^ 2 sin <(t)> & amp 0 & amp- cos <(t)> & amp -t ^ 2 end] e o forçamento é [ fBld (t) = begin 0 0 0 te ^ fim.]

Exercício 8

Considere a equação de segunda ordem (y & # 39 & # 39 + p (t) y & # 39 + q (t) y = g (t) ) com condições iniciais (y & # 39 (0) = 1 ) e (y (0) = 2 ). Deixe (x_1 = y ) e (x_2 = y & # 39 ), e então expresse esta equação de segunda ordem como um sistema de duas equações de primeira ordem. Certifique-se de incluir a condição inicial do seu sistema.

Solução

Como um sistema, teríamos [ begin x_1 & # 39 & amp = x_2 x_2 & # 39 & amp = -q (t) x_1-p (t) x_2 + g (t) fim] com condições iniciais (x_1 (0) = 2 ) e (x_2 (0) = 1 ). Como uma equação vetorial, pode ser escrita como [ xBld ^ prime = begin 0 & amp 1 -q (t) & amp -p (t) end < bf x> + begin 0 g (t) fim, ] com condição inicial ( xBld (0) = begin 2 1 fim) .

Exercício 9

Considere a equação de enésima ordem (y ^ <(n)> + a_1 (t) y ^ <(n-1)> + a_2 (t) y ^ <(n-2)> + pontos + a_(t) y & # 39 + a_n (t) y = g (t) ) com condições iniciais (y ^ <(i)> (0) = b_) para (i = 0, pontos, n-1 ). Expresse esta equação de enésima ordem como um sistema de (n ) equações de primeira ordem. Certifique-se de incluir a condição inicial do seu sistema.

Solução

Conforme mostrado no texto, se definirmos [ xBld = begin x_1 x_2 vdots x_n end = begin y y ^ prime vdots y ^ <(n-1)>, end] o sistema correspondente é então [ xBld ^ prime = begin 0 & amp 1 & amp 0 & amp cdots & amp 0 0 & amp ddots & amp ddots & amp ddots & amp vdots vdots & amp ddots & amp 0 & amp 1 & amp 0 0 & amp cdots & amp 0 & amp 0 & amp 1 -a_n (t) & amp cdots & amp -a_3 (t) & amp a_2 (t) e amp -a_1 (t) end < bf x> + begin 0 0 vdots 0 g (t) end, quad text xBld (0) = begin b_1 b_2 vdots b_n end.]

Exercício 10

etiqueta Dois pêndulos suspensos de comprimento ( ell ) e massa (m_1 ) e (m_2 ) são acoplados por uma mola. Sejam ( theta_1 ) e ( theta_2 ) o deslocamento angular de cada pêndulo de sua posição de repouso. Para ângulos pequenos, as equações de movimento são aproximadas pelo seguinte sistema linear [ begin m_1 ell theta_1 ^ < prime prime> & amp = - m_1 g theta_1 - k ell ( theta_1 - theta_2) m_2 ell theta_2 ^ < prime prime> & amp = - m_2 g theta_2 + k ell ( theta_1 - theta_2). fim] Escreva isso como um sistema linear de primeira ordem. Encontre a matriz de coeficiente correspondente ( ABld ) e o forçamento ( fBld ).

Solução

Escreva as equações na forma normal [ begin theta_1 ^ < prime prime> & amp = - frac < ell> theta_1 - frac( theta_1 - theta_2) theta_2 ^ < prime prime> & amp = - frac < ell> theta_2 + frac( theta_1 - theta_2). fim] Defina [ xBld = begin x_1 x_2 x_3 x_4 end = begin theta_1 theta_1 ^ prime theta_2 theta_2 ^ prime end] Pegando a derivada em ambos os lados e usando os EDOs, encontramos [ xBld ^ prime = begin x_2 - frac< ell> x_1 - frac(x_1 - x_3) x_4 - frac < ell> x_3 + frac(x_1 - x_3) fim= begin 0 & amp 1 & amp 0 & amp 0 - frac < ell> - frac & amp 0 & amp frac & amp 0 0 & amp 0 & amp 0 & amp 1 frac & amp 0 & amp - frac < ell> - frac & amp 0 end xBld. ]

Para os problemas 11-12, determine o maior intervalo onde existe uma solução única para os seguintes problemas de valor inicial para sistemas de primeira ordem.

Exercício 11

Solução

Observe que os coeficientes não são definidos quando (t leq-1 ), (t & gt3 ), (t = (n + frac <1> <2>) pi / 2 ) e (t = 5 ). O maior intervalo que contém o tempo inicial (t = 0 ) onde todas as funções são contínuas é, portanto, ((- 1, pi / 2) ).

Exercício 12

Solução

Os coeficientes e forçantes deste sistema não são definidos sempre que (t & lt0 ) ou quando (t = 2 ). Portanto, o maior intervalo contendo o tempo inicial (t = 1 ) é (0,2) ).

Exercício 13

Considere o sistema diferencial [ frac < dee> < dt> begin x y end = begin 3 & amp 1 2 & amp 2 end começar x y end]

Mostre que ( begine ^ <4t> e ^ <4t> end) e ( begin-e ^ 2e ^fim) são ambas as soluções para este sistema.

Dê uma matriz fundamental para este sistema.

Dê uma solução geral para este sistema na forma vetorial.

Calcule a matriz fundamental natural para este sistema associado a (t = 0 ).

Resolva o problema do valor inicial para este sistema com (x (0) = -2 ) e (y (0) = 3 ).

Solução

Essas soluções formam um conjunto fundamental, já que o Wronskian é [ det begin e ^ <4t> & amp -e ^ t e ^ <4t> & amp 2e ^ fim = 3e ^ <5t> neq 0. ] Uma matriz fundamental para este sistema é, portanto, dada por [ PsiBld (t) = begin e ^ <4t> & amp -e ^ t e ^ <4t> & amp 2e ^ fim.]

Uma solução geral é dada por [ xBld (t) = PsiBld (t) cBld = c_1 begine ^ <4t> e ^ <4t> end + c_2 begin -e ^ 2e ^fim.]

Podemos obter a matriz natural fundamental ( PhiBld (t) ) dada nossa matriz fundamental ( PsiBld (t) ) por ( PhiBld (t) = PsiBld (t) PsiBld (0) ^ < -1> ). Obtemos [ PhiBld (t) = begin e ^ <4t> & amp -e ^ t e ^ <4t> & amp 2e ^ t end frac <1> <3> begin 2 & amp 1 -1 & amp 1 end = frac <1> <3> begin 2e ^ <4t> + e ^ t & amp e ^ <4t> - e ^ t 2e ^ <4t> -2 e ^ t & amp e ^ <4t> + 2e ^ t end.]

A solução para o problema do valor inicial é [ xBld (t) = PhiBld (t) xBld ^ I = frac <1> <3> begin 2e ^ <4t> + e ^ t & amp e ^ <4t> - e ^ t 2e ^ <4t> -2 e ^ t & amp e ^ <4t> + 2e ^ t end começar -2 3 fim = frac <1> <3> begin -e ^ <4t> -5e ^ t -e ^ <4t> + 10e ^ t end]

Exercício 14

Considere o sistema diferencial [ frac < dee> < dt> begin x y end= begin 1 & amp - cos <(t)> cos <(t)> & amp 1 end começar x y end]

Mostre que ambos ( começame ^ t cos ( sin (t)) e ^ t sin ( sin (t)) end) e ( begin-e ^ t sin ( sin (t)) e ^ t cos ( sin (t)) end) são ambas as soluções para este sistema.

Dê uma matriz fundamental para este sistema.

Escreva uma solução geral para este sistema na forma vetorial.

Calcule a matriz fundamental natural deste sistema associado a (t = 0 ).

Resolva o problema do valor inicial para este sistema com (x (0) = 1 ), (y (0) = 0 ).

Solução

Essas soluções formam um conjunto fundamental, uma vez que o Wronskian é [ det begin e ^ t cos <( sin <(t)>)> & amp -e ^ t sin <( sin <(t)>)> e ^ t sin <( sin <(t)> )> & amp e ^ t cos <( sin <(t)>)> end = e ^ <2t> neq 0. ] Uma matriz fundamental para este sistema é, portanto, dada por [ PsiBld (t) = begin e ^ t cos <( sin <(t)>)> & amp -e ^ t sin <( sin <(t)>)> e ^ t sin <( sin <(t)> )> & amp e ^ t cos <( sin <(t)>)> end.]

Uma vez que ( PsiBld (0) = IBld ) nossa matriz fundamental dada na parte (b) é na verdade a matriz fundamental natural [ PhiBld (t) = PsiBld (t) = begin e ^ t cos <( sin <(t)>)> & amp -e ^ t sin <( sin <(t)>)> e ^ t sin <( sin <(t)> )> & amp e ^ t cos <( sin <(t)>)> end.]

Exercício 15

Considere o sistema diferencial [ frac < dee> < dt> begin x y end = begin 2t & amp -e ^ e ^ <-t ^ 2> & amp 0 end começar x y end]

Mostre que ambos ( começame ^ cos <(t)> sin <(t)> end) e ( begine ^ sin <(t)> - cos <(t)> end) são soluções para o sistema acima.

Dê uma matriz fundamental para este sistema.

Escreva uma solução geral para este sistema na forma vetorial.

Calcule a matriz fundamental natural deste sistema associado a (t = 0 ).

Resolva o problema do valor inicial para este sistema com (x (0) = -1 ), (y (0) = 3 ).

Solução

Podemos ver que essas soluções formam um conjunto fundamental, já que o Wronskian é [ det begin e ^ cos <(t)> & amp e ^ sin <(t)> sin <(t)> & amp - cos <(t)> end = -e ^ neq 0. ] Portanto, uma matriz fundamental para este sistema é dada por [ PsiBld (t) = begin e ^ cos <(t)> & amp e ^ sin <(t)> sin <(t)> & amp - cos <(t)> end.]

Uma solução geral é [ xBld (t) = PsiBld (t) cBld = c_1 begin e ^ cos <(t)> sin <(t)> end + c_2 begin e ^ sin <(t)> - cos <(t)> end]

Podemos obter a matriz natural fundamental ( PhiBld (t) ) dada nossa matriz fundamental ( PsiBld (t) ) por ( PhiBld (t) = PsiBld (t) PsiBld (0) ^ < -1> ). Obtemos [ PhiBld (t) = begin e ^ cos <(t)> & amp e ^ sin <(t)> sin <(t)> & amp - cos <(t)> end começar 1 & amp 0 0 & amp -1 end= begin e ^ cos <(t)> & amp -e ^ sin <(t)> sin <(t)> & amp cos <(t)> end.]

A solução para o problema do valor inicial é [ xBld (t) = PhiBld (t) xBld ^ I = begin e ^ cos <(t)> & amp -e ^ sin <(t)> sin <(t)> & amp cos <(t)> end começar -1 3 fim = begin -e ^ cos <(t)> - 3e ^ sin <(t)> 3 cos <(t)> - sin <(t)> end]

Exercício 16

Considere o sistema diferencial [ frac < dee> < dt> begin x y end = frac <1> <1 + t ^ 2> begin (1 + t) ^ 2 e amp 2 (1 + t ^ 2) ^ 2 2 e amp 1 + t ^ 2 end começar x y end]

Mostre que ambos ( começam(1 + t ^ 2) e ^ <3t> e ^ <3t> end) e ( begin(1 + t ^ 2) e ^ <-t> -e ^ <-t> end) são soluções para o sistema acima.

Dê uma matriz fundamental para este sistema.

Escreva uma solução geral para este sistema na forma vetorial.

Calcule a matriz fundamental natural deste sistema associado a (t = 0 ).

Resolva o problema do valor inicial para este sistema com (x (0) = -4 ), (y (0) = 2 ).

Solução

Estas soluções formam um conjunto fundamental desde [ det begin (1 + t ^ 2) e ^ <3t> & amp (1 + t ^ 2) e ^ <-t> e ^ <3t> & amp -e ^ <-t> end] Portanto, uma matriz fundamental para este sistema é dada por [ PsiBld (t) = begin (1 + t ^ 2) e ^ <3t> & amp (1 + t ^ 2) e ^ <-t> e ^ <3t> & amp -e ^ <-t> end.]

Uma solução geral é [ xBld (t) = PsiBld cBld = c_1 begin (1 + t ^ 2) e ^ <3t> e ^ <3t> end + c_2 begin (1 + t ^ 2) e ^ <-t> e ^ <-t> end.]

Podemos obter a matriz natural fundamental ( PhiBld (t) ) de nossa matriz fundamental ( PsiBld (t) ), pela fórmula ( PhiBld (t) = PsiBld (t) PsiBld (0 ) ^ <-1> ). Encontramos [ begin PhiBld (t) & amp = begin (1 + t ^ 2) e ^ <3t> & amp (1 + t ^ 2) e ^ <-t> e ^ <3t> & amp e ^ <-t> end começar 1 & amp 1 1 & amp -1 end & amp = begin (1 + t ^ 2) (e ^ <3t> + e ^ <-t>) & amp (1 + t ^ 2) (e ^ <3t> - e ^ <-t>) e ^ <3t> + e ^ <-t> & amp e ^ <3t> - e ^ <-t> end fim]

Para os problemas 17-20, mostre que as seguintes soluções com valores vetoriais para um sistema linear formam um conjunto fundamental. Encontre o sistema linear que eles resolvem.

Exercício 17

Solução

Vemos que essas soluções formam um conjunto fundamental, uma vez que seu Wronskian [W (t) = det begin cos <(t)> & amp - sin <(t)> sin <(t)> + cos <(t)> & amp cos <(t)> - sin <(t)> fim= 1 neq 0. ] Para encontrar o sistema linear que este conjunto fundamental resolve, usamos uma matriz fundamental [ PsiBld (t) = begin cos <(t)> & amp - sin <(t)> sin <(t)> + cos <(t)> & amp cos <(t)> - sin <(t)> fim, ] e, portanto, a matriz de coeficientes ( ABld (t) ) é dada por [ begin ABld (t) & amp = PsiBld ^ < prime> (t) PsiBld ^ <-1> (t) & amp = begin - sin <(t)> & amp - cos <(t)> cos <(t)> - sin <(t)> & amp - sin <(t)> - cos <(t) > end começar cos <(t)> - sin <(t)> & amp sin <(t)> - sin <(t)> - cos <(t)> & amp cos <(t)> fim & amp = begin 1 e amp -1 2 e amp -1 end fim]

Exercício 18

Solução

Vemos que essas soluções formam um conjunto fundamental, uma vez que seu Wronskian [W (t) = det begin e ^ t & amp -e ^ <-2t> e ^ t & amp 3e ^ <-2t> end= 4e ^ <-t> neq 0. ] Para encontrar o sistema linear que este conjunto fundamental resolve, usamos uma matriz fundamental [ PsiBld (t) = begin e ^ t & amp -e ^ <-2t> e ^ t & amp 3e ^ <-2t> end] e, portanto, a matriz de coeficientes ( ABld (t) ) é dada por [ begin ABld (t) & amp = PsiBld ^ < prime> (t) PsiBld ^ <-1> (t) & amp = begin e ^ t & amp 2e ^ <-2t> e ^ t & amp -6e ^ <-2t> end frac <1> <4> begin 3e ^ <-t> & amp e ^ <-t> -e ^ <2t> & amp e ^ <2t> end & amp = frac <1> <4> begin 1 & amp 3 9 & amp -5 end. fim]

Exercício 19

Solução

Vemos que essas soluções formam um conjunto fundamental, pois seu Wronskian [W (t) = det begin 2te ^ <2t> & amp -2e ^ <-3t> 1-te ^ <2t> & amp e ^ <-3t> end= 2e ^ <-3t> neq 0. ] Para encontrar o sistema linear que este conjunto fundamental resolve, usamos uma matriz fundamental [ PsiBld (t) = begin 2te ^ <2t> & amp -2e ^ <-3t> 1-te ^ <2t> & amp e ^ <-3t> end] e, portanto, a matriz de coeficientes ( ABld (t) ) é dada por [ begin ABld (t) & amp = PsiBld ^ < prime> (t) PsiBld ^ <-1> (t) & amp = begin 2e ^ <2t> + 4te ^ <2t> & amp 6e ^ <-3t> -e ^ <2t> - 2te ^ <2t> & amp -3e ^ <-3t> end frac <1> <2> begin 1 & amp 2 te ^ <5t> -1e ^ <3t> & amp 2te ^ <5t> end & amp = begin (1 + 5t) e ^ <2t> - 3 & amp (2 + 10t) e ^ <2t> - frac <1> <2> (1 + 5t) e ^ <2t> + frac <1> <2> & amp - (1 + 5t) e <2t> end. fim]

Exercício 20

Solução

Vemos que essas soluções formam um conjunto fundamental desde seu Wronskian [ begin W (t) & amp = det begin 1 & amp t + 1 & amp frac <1> <2> t ^ 2 + t +1 0 & amp 1 & amp t + 1 2 & amp 2t + 2 & amp (t + 1) ^ 2 ende ^ <3t> & amp = e ^ <9t> ((t + 1) ^ 2 + 2 (t + 1) ^ 2 - t ^ 2-2t -2 - 2 (t + 1) ^ 2) & amp = -e ^ <9t> neq 0 end] Para encontrar o sistema linear que este conjunto fundamental resolve, usamos uma matriz fundamental [ PsiBld (t) = begin 1 & amp t + 1 & amp frac <1> <2> t ^ 2 + t +1 0 & amp 1 & amp t + 1 2 & amp 2t + 2 & amp (t + 1) ^ 2 ende ^ <3t> ] e, portanto, a matriz de coeficiente ( ABld (t) ) é dada por ( ABld (t) = PsiBld ^ prime (t) PsiBld ^ <-1> (t) ). Pode-se calcular o inverso como [ PsiBld ^ <-1> (t) = begin (t + 1) ^ 2 & amp -t-1 & amp - frac <1> <2> t ^ 2 -t -2t -2 & amp 1 & amp t + 1 2 & amp 0 & amp -1 ende ^ <-3t> ] e a derivada [ PsiBld ^ < prime> (t) = begin 3 & amp 3t + 4 & amp frac <3> <2> t ^ 2 + 4t + 4 0 & amp 3 & amp 3t + 4 6 & amp 6t +8 & amp 3t ^ 2 + 8t + 5 end. ] Portanto, após algum trabalho, [ começar ABld (t) & amp = begin 3 & amp 3t + 4 & amp frac <3> <2> t ^ 2 + 4t + 4 0 & amp 3 & amp 3t + 4 6 & amp 6t +8 & amp 3t ^ 2 + 8t + 5 end começar 3 & amp 3t + 4 & amp frac <3> <2> t ^ 2 + 4t + 4 0 & amp 3 & amp 3t + 4 6 & amp 6t +8 & amp 3t ^ 2 + 8t + 5 end & amp = begin 3 & amp 1 & amp 0 2 & amp 3 & amp -1 0 & amp 2 & amp 3 end. fim]

Exercício 21

Seja ( PhiBld (t) ) a matriz natural fundamental associada a (t_I ) para o sistema [ frac < dee xBld> < dt> = ABld xBld, ] onde ( ABld ) é uma matriz constante. Mostre que ( PhiBld (t) ) e ( ABld ) comutam, ou seja, [ PhiBld (t) ABld = ABld PhiBld (t). ] Dica: Mostre que ( PhiBld (t) ABld ) mostra que ( ABld PhiBld (t) ) resolve o mesmo problema de valor inicial.

Solução

Tomando as derivadas de tempo de ( PhiBld (t) ABld ) e ( ABld PhiBld (t) ), encontramos [ begin ( PhiBld ABld) ^ prime & amp = PhiBld ^ prime ABld = ABld ( PhiBld ABld) ( ABld PhiBld) ^ prime & amp = ABld PhiBld ^ prime = ABld ( ABld PhiBld). fim] No entanto, por definição ( PhiBld (t_I) = IBld ), e assim ( PhiBld (t_I) ABld = ABld PhiBld (t_I) = ABld ). Portanto, pela unicidade do problema inicial com valor de matriz, concluímos que [ ABld PhiBld (t) = PhiBld (t) ABld. ]

Exercício 22

Seja ( PsiBld (t) ) uma matriz fundamental para o sistema [ frac < dee xBld> < dt> = ABld (t) xBld. ] Para qualquer matriz constante ( CBld ) de forma que ( det ( CBld) neq 0 ), mostre que ( PsiBld CBld ) também é uma matriz fundamental, mas que ( CBld PsiBld ) pode não ser. Como deve ( CBld ) e ( ABld (t) ) estar relacionados para que ( CBld PsiBld ) se torne uma matriz fundamental?

Solução

Uma matriz fundamental ( PsiBld ) deve satisfazer a equação diferencial [ notag PsiBld ^ prime = ABld (t) PsiBld ] e ter ( det <( PsiBld)> neq 0 ) . Ao multiplicar a equação acima à direita em ambos os lados por ( CBld ), encontramos [ notag ( PsiBld CBld) ^ < prime> = ABld (t) ( PsiBld CBld), ] e usando as propriedades do determinante ( det ( PsiBld CBld) = det <( PsiBld)> det <( CBld)> neq 0 ). Portanto, ( PsiBld CBld ) também é uma matriz fundamental. No entanto, tomando a derivada de ( CBld PsiBld ), encontramos [ notag ( CBld PsiBld) ^ prime = CBld ABld (t) PsiBld neq ABld (t) ( CBld PsiBld). ] Podemos ver então que se ( ABld (t) ) e ( CBld ) comutarem, isto é ( ABld (t) CBld = CBld ABld (t) ) , então ( CBld PsiBld ) será uma matriz fundamental.

Exercício 23

Sejam ( PsiBld_1 ) e ( PsiBld_2 ) duas matrizes fundamentais de [ frac < dee xBld> < dt> = ABld (t) xBld. ] Mostre que existe uma constante ( CBld ), ( det <( CBld)> neq 0 ), de modo que ( PsiBld_1 = PsiBld_2 CBld ).
Dica: mostre que (( PsiBld_2 ^ <-1> PsiBld_1) ^ prime = ZeroBld ) e use a regra do produto para matrizes (( ABld BBld) ^ prime = ABld ^ prime BBld + ABld BBld ^ prime ) (veja os exercícios no suplemento sobre matrizes e vetores).

Solução

Vemos por computação direta que [ notag begin ABld PsiBld_1 = PsiBld_1 ^ prime & amp = ( PsiBld_2 PsiBld_2 ^ <-1> PsiBld_1) ^ prime & amp = PsiBld_2 ^ prime ( PsiBld_2 ^ <-1> PsiBld_1) + PsiBld_2 ( PsiBld_2 ^ <-1> PsiBld_1) ^ prime & amp = ABld PsiBld_2 PsiBld_2 ^ <-1> PsiBld_1 + PsiBld_2 ( PsiBld_2 ^ <-1> PsiBld_1) ^ prime & amp = ABld PsiBld_1 + PsiBld_2 ( PsiBld_2 ^ <-1> PsiBld_1) ^ prime. fim] Subtraindo ( ABld PsiBld_1 ) de ambos os lados, obtemos [ notag PsiBld_2 ( PsiBld_2 ^ <-1> PsiBld_1) ^ prime = ZeroBld. ] Desde ( PsiBld_2 ) é invertível, podemos concluir [ notag ( PsiBld_2 ^ <-1> PsiBld_1) ^ prime = ZeroBld. ] e, portanto, há uma matriz constante ( CBld ) tal que ( PsiBld_2 ^ <-1> PsiBld_1 = CBld ) e [ notag det <( CBld)> = frac < det <( PsiBld_1) >> < det <( PsiBld_2) >> neq 0. ]

Exercício 24

O problema anterior implica que se ( PsiBld_1 ) e ( PsiBld_2 ) são matrizes fundamentais de [ frac < dee xBld> < dt> = ABld (t) xBld, ] então seus Wronskians (W_1 (t) = det <( PsiBld_1 (t))> ), (W_2 (t) = det <( PsiBld_2 (t))> ) diferem por um múltiplo constante (c neq 0 ), uma vez que [W_1 (t) = det <( PsiBld_1 (t))> = det <( CBld)> det <( PsiBld_2 (t))> = cW_2 (t) . ] Use o Teorema Wronskiano de Liouville para chegar à mesma conclusão.

Solução

O Teorema Wronskiano de Liouville afirma que dada qualquer solução fundamental ( PsiBld (t) ) para [ frac < dee xBld> < dt> = ABld (t) xBld ], então o Wronskian (W ( t) = det <( PsiBld (t))> ) deve satisfazer [ frac < dee> < dt> W (t) = tr left ( ABld (t) right) W ( t). ] Sejam ( PsiBld_1 (t) ) e ( PsiBld_2 (t) ) duas soluções fundamentais para o sistema acima, e sejam (W_1 (t) ), (W_2 (t) ) ) ser seus Wronskians. Mostraremos que (W_1 (t) / W_2 (t) ) é constante. Na verdade, tomando sua derivada e usando o Teorema Wronskiano de Liouville, encontramos [ begin left ( frac right) ^ prime & amp = frac - frac & amp = tr left ( ABld right) left ( frac - frac direita) = 0. fim] Portanto, concluímos que existe um (c neq 0 ) tal que (W_1 (t) = cW_2 (t) ).

Exercício 25

Considere a equação diferencial linear de ordem homogênea (n ^ < mathrm th> ) na forma normal [ frac < dee ^ n y> < dt> + a_1 (t) frac < dee ^y> < dt> + ldots + a_(t) frac < dee y> < dt> + a_n (t) y = 0, ] e seu sistema linear de primeira ordem (n ) -dimensional equivalente [ frac < dee xBld> < dt> = ABld (t) xBld. ] Mostre o Teorema Wronskiano de Abel para a equação de ordem (n ^ < mathrm th> ) usando o Teorema Wronskiano de Liouville para o sistema linear de primeira ordem.

Solução

O sistema linear de ordem (n ^ < mathrm th> ) pode ser escrito como um sistema linear de primeira ordem (n ) -dimensional [ frac < dee xBld> < dt> = ABld (t ) xBld, qquad text xBld = begin x_1 x_2 vdots x_n end = begin y y ^ prime vdots y ^ <(n-1)> end, ] e [ ABld (t) = begin 0 & amp 1 & amp 0 & amp cdots & amp 0 0 & amp ddots & amp ddots & amp ddots & amp vdots vdots & amp ddots & amp 0 & amp 1 & amp 0 0 & amp cdots & amp 0 & amp 0 & amp 1 -a_n (t) & amp cdots & amp -a_3 (t) & amp a_2 (t) & amp -a_1 (t) end] Se (Y_1, Y_2, ldots Y_n ) são (n ) soluções linearmente independentes para o (n ^ < mathrm th> ) sistema linear de ordem, então, correspondentemente [ xBld_1 = begin Y_1 Y_1 ^ prime vdots Y ^ <(n-1)> _ 1 fim, quad xBld_2 = begin Y_2 Y_2 ^ prime vdots Y ^ <(n-1)> _ 2 end, quad ldots xBld_n = begin Y_n Y_n ^ prime vdots Y ^ <(n-1)> _ n fim] são um conjunto fundamental de soluções. Além disso, os Wronskians definem a mesma quantidade [W [Y_1, Y_2, ldots, Y_n] (t) = W [ xBld_1, xBld_2, ldots, xBld_n] (t). ] Portanto, desde ( tr left ( ABld (t) right) = -a_1 (t) ), Wronskian de Liouvilles imediatamente implica [ frac < dee> < dt> W [Y_1, Y_2, ldots Y_n] (t) = -a_1 (t) W [Y_1, Y_2, ldots Y_n] (t), ] que é apenas o Teorema Wronskiano de Abel.


250+ TOP MCQs em equações diferenciais lineares de primeira ordem | Aula 12 de matemática

Questões de múltipla escolha em matemática para exames de admissão em engenharia sobre “Equações diferenciais lineares de primeira ordem - 2”.

1. Uma curva passa por (1, 1) de forma que o triângulo formado pelos eixos coordenados e a tangente em qualquer ponto da curva está no primeiro quadrante e tem sua área igual a 2. Qual é a equação diferencial?
a) dy / dx = [(xy + 2) ± √ (1 + xy)] / x 2
b) dy / dx = [(xy - 2) ± √ (1 + xy)] / x 2
c) dy / dx = [(xy - 2) ± √ (1 - xy)] / x 2
d) dy / dx = [(xy + 2) ± √ (1 - xy)] / x 2
Resposta: c
Esclarecimento: A equação da tangente à curva y = f (x), no ponto (x, y), é
Y - y = dy / dx * (X - x)… .. (1)
Onde encontra o eixo x, Y = 0 e X = (x - y / (dy / dx))
Onde encontra o eixo y, X = 0 e Y = (y - x / (dy / dx))
Além disso, a área do triângulo formado por (1) com os eixos coordenados é 2, de modo que,
(x - y / (dy / dx)) * (y - x / (dy / dx)) = 4
Ou, (y - x / (dy / dx)) 2 - 4dy / dx = 0
Ou, x 2 (dy / dx) 2 - 2 (xy - 2) dy / dx + y 2 = 0
Resolvendo para dy / dx, obtemos,
dy / dx = [(xy - 2) ± √ (1 - xy)] / x 2

2. Uma curva passa por (1, 1) de forma que o triângulo formado pelos eixos coordenados e pela tangente em qualquer ponto da curva esteja no primeiro quadrante e tenha sua área igual a 2. Qual será a equação da curva ?
a) xy = 2
b) xy = -1
c) x - y = 2
d) x + y = 2
Resposta: d
Esclarecimento: A equação da tangente à curva y = f (x), no ponto (x, y), é
Y - y = dy / dx * (X - x) ... .. (1)
Onde encontra o eixo x, Y = 0 e X = (x - y / (dy / dx))
Onde encontra o eixo y, X = 0 e Y = (y - x / (dy / dx))
Além disso, a área do triângulo formado por (1) com os eixos coordenados é 2, de modo que,
(x - y / (dy / dx)) * (y - x / (dy / dx)) = 4
Ou, (y - x / (dy / dx)) 2 - 4dy / dx = 0
Ou, x 2 (dy / dx) 2 - 2 (xy - 2) dy / dx + y 2 = 0
Resolvendo para dy / dx, obtemos,
dy / dx = [(xy - 2) ± √ (1 - xy)] / x 2
Deixe, 1 - xy = t 2
=> x (dy / dx) + y = -2t (dt / dx)
=> x 2 (dy / dx) = t 2 - 1 - 2tx (dt / dx), de modo que (3) dá
t (x (dt / dx) - (t ± 1)) = 0
Portanto, t = 0
=> xy = 1 que é satisfeito por (1, 1)
Ou, x dt / dx = t ± 1
=> dx / x = dt / t ± 1
=> t ± 1 = cx
Para x = 1, y = 1 e t = 0
=> c = ± 1, então a solução é
t = ± (x - 1) => t 2 = (x - 1) 2
Ou, 1 - xy = x 2 - 2x + 1
Ou, x + y = 2
Assim, as duas curvas que satisfazem são xy = 1 e x + y = 2

3. Qual será o valor de dy / dx - a / x * y = (x + 1) / x?
a) y = x / (1 - a) - 1 / a + cx a
b) y = x / (1 + a) + 1 / a + cx a
c) y = x / (1 - a) - 1 / a - cx a
d) y = x / (1 + a) - 1 / a + cx a
Resposta: a
Esclarecimento: dy / dx - a / x * y = (x + 1) / x ……. (1)
Multiplicando ambos os lados da equação (1) por
e ∫-a / xdx
= e -a log x
= e log x -a
= x -a
Obtemos, x -a dy / dx - x -a (a / x) y = x -a (x + 1) / x
Ou, d / dx (y. X -a) = x -a + x -a - 1 ……. (2)
Integrando ambos os lados de (2), obtemos,
y. x -a = x -a + 1 / (- a + 1) + x -a - 1 + 1 / (- a -1 + 1) + c
= x -a .x / (1 - a) + x -a / -a + c
Ou, y = x / (1 - a) - 1 / a + cx a

4. Qual será a forma da equação diferencial de √ (a 2 + x 2) dy / dx + y = √ (a 2 + x 2) - x?
a) a 2 log (x + √ (a 2 - x 2)) + c
b) a 2 log (x + √ (a 2 + x 2)) + c
c) a 2 log (x - √ (a 2 + x 2)) + c
d) a 2 log (x - √ (a 2 - x 2)) + c
Resposta: b
Esclarecimento: A forma dada de equação pode ser escrita como,
dy / dx + 1 / √ (a 2 + x 2) * y = (√ (a 2 + x 2) - x) / √ (a 2 + x 2) …… (1)
Temos, ∫1 / √ (a 2 + x 2) dx = log (x + √ (a 2 + x 2))
Portanto, o fator de integração é,
e ∫1 / √ (a 2 + x 2) = e log (x + √ (a 2 + x 2))
= x + √ (a 2 + x 2)
Portanto, multiplicando ambos os lados de (1) por x + √ (a 2 + x 2) obtemos,
x + √ (a 2 + x 2 dy / dx + (x + √ (a 2 + x 2)) / √ (a 2 + x 2) * y = (x + √ (a 2 + x 2)) ( √ (a 2 + x 2) - x) / √ (a 2 + x 2)
ou, d / dx [x + √ (a 2 + x 2) * y] = (a 2 + x 2) ……… .. (2)
Integrando ambos os lados de (2), obtemos,
(x + √ (a 2 + x 2) * y = a 2 ∫dx / √ (a 2 + x 2)
= a 2 log (x + √ (a 2 + x 2)) + c

5. Qual é a solução de dy / dx = (6x + 9y - 7) / (2x + 3y - 6)?
a) 3x - y + log | 2x + 3y - 3 | = -c / 3
b) 3x - y + log | 2x + 3y - 3 | = c / 3
c) 3x + y + log | 2x + 3y - 3 | = -c / 3
d) 3x - y - log | 2x + 3y - 3 | = c / 3
Resposta: a
Esclarecimento: dy / dx = (6x + 9y - 7) / (2x + 3y - 6)
Portanto, dy / dx = (3 (2x + 3y) - 7) / (2x + 3x - 6) ………. (1)
Agora, colocamos 2x + 3y = z
Portanto, 2 + 3dy / dx = dz / dx [diferenciando em relação a x]
Ou dy / dx = 1/3 (dz / dx - 2)
Portanto, de (1) obtemos,
1/3 (dz / dx - 2) = (3z - 7) / (z - 6)
Ou dz / dx = 2 + (3 (3z - 7)) / (z - 6)
= 11 (z - 3) / (z - 6)
Ou, (z - 6) / (z - 3) dz = 11 dx
Ou, ∫ (z - 6) / (z - 3) dz = ∫11 dx
Ou, ∫ (1 - 3 / (z - 3)) dz = 11x + c
Ou, z - log | z - 3 | = 11x + c
Ou 2x + 3a - 11x - 3log | 2x + 3a -3 | = c
Ou 3a - 9x - 3log | 2x + 3a - 3 | = c
Ou 3x - y + log | 2x + 3y - 3 | = -c / 3

6. Uma partícula parte da origem com uma velocidade de 5cm / seg e move-se em linha reta, sendo sua aceleração no tempo t segundos (3t 2 - 5t) cm / seg 2. Qual será a velocidade da partícula?
a) 27cm / seg
b) 28 cm / s
c) 29 cm / s
d) 30 cm / s
Resposta: c
Esclarecimento: Seja x cm a distância da partícula em movimento da origem ev cm / s sua velocidade no final de t segundos. Então, a aceleração da partícula no tempo t segundos = dv / dt e sua velocidade naquele tempo = v = dx / dt.
Por questão, dv / dt = 3t 2 - 5t
Ou dv = 3t 2 dt - 5tdt
Ou, ∫dv = 3∫t 2 dt - 5∫t dt
Ou, v = t 3 - (5/2) t 2 + c ………. (1)
Dado, v = 5, quando t = 0, portanto, colocando esses valores na equação (1) obtemos, c = 5
Assim, v = t 3 - (5/2) t 2 + 5
Ou, dx / dt = t 3 - (5/2) t 2 + 5 ……… .. (2)
Assim, a velocidade da partícula ao final de 4 segundos,
= [v]t = 4 = (4 3 – (5/2)4 2 + 5 ) cm/sec [putting t = 4 in (2)]
= 29 cm/sec

7. A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t 2 – 5t)cm/sec 2 . What will be the distance from the origin at the end of 4 seconds?
a) 30(4/3)
b) 30(2/3)
c) 30
d) Unpredictable
Answer: b
Clarification: Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.
By question, dv/dt = 3t 2 – 5t
Or, dv = 3t 2 dt – 5t dt
Or, ∫dv = 3∫t 2 dt – 5∫t dt
Or, v = t 3 – (5/2)t 2 + c ……….(1)
Given, v = 5, when t = 0 hence putting these values in equation (1) we get, c = 5
Thus v = t 3 – (5/2)t 2 + 5
Or, dx/dt = t 3 – (5/2)t 2 + 5 ………..(2)
Or, dx = t 3 dt – (5/2)t 2 dt + 5 dt
Integrating this we get,
x = (1/4)t 4 – (5/2)t 3 /3 + 5t + k ……….(3)
By the problem, x = 0, when t = 0 hence, from (3) we get, k = 0.
Thus, x = (1/4)t 4 – (5/6)t 3 + 5t ……….(4)
Thus, the velocity of the particle at the end of 4 seconds,
= [x]t = 4 = (1/4)4 4 – (5/6)4 3 + 5(4) [putting t = 4 in (4)]
= 30(2/3) cm

8. What is the solution of (y(dy/dx) + 2x) 2 = (y 2 + 2x 2 )[1 + (dy/dx) 2 ]?
a) cx ±1/√2 = y/x + √(y 2 – 2x 2 )/x 2
b) cx ±√2 = y/x + √(y 2 + 2x 2 )/x 2
c) cx ±1/2√2 = y/x + √(y 2 – 2x 2 )/x 2
d) cx ±1/√2 = y/x + √(y 2 + 2x 2 )/x 2
Answer: d
Clarification: Here, y 2 (dy/dx) 2 + 4x 2 + 4xy(dy/dx) = (y 2 + 2x 2 )[1 + (dy/dx) 2 ]
=>dy/dx = y/x ± √(1/2(y/x) 2 ) + 1
Let, y = vx
=> v + x dv/dx = v ± √(1/2(v) 2 ) + 1
Integrating both sides,
±√dv/(√(1/2(v) 2 ) + 1) = ∫dx/x
cx ±1/√2 = y/x + √(y 2 + 2x 2 )/x 2

9. What is the equation of the curve passing through (1, 0) of (y(dy/dx) + 2x) 2 = (y 2 + 2x 2 )[1 + (dy/dx) 2 ]?
a) √2x ±1/√2 = y/x + √(y 2 + 2x 2 )/x 2
b) √2x ±1/2√2 = y/x + √(y 2 + 2x 2 )/x 2
c) √2x √2 = y/x + √(y 2 + 2x 2 )/x 2
d) √2x = y/x + √(y 2 + 2x 2 )/x 2
Answer: a
Clarification: Here, y 2 (dy/dx) 2 + 4x 2 + 4xy(dy/dx) = (y 2 + 2x 2 )[1 + (dy/dx) 2 ]
=> dy/dx = y/x ± √(1/2(y/x) 2 ) + 1
Let, y = vx
=> v + x dv/dx = v ± √(1/2(v) 2 ) + 1
Integrating both sides,
±∫dv/(√(1/2(v) 2 ) + 1) = ∫dx/x
cx ±1/√2 = y/x + √(y 2 + 2x 2 )/x 2 (put v/√2 = tan t)
putting x = 1, y = 0, we get c = √2
So, the curve is given by,
√2x ±1/√2 = y/x + √(y 2 + 2x 2 )/x 2

10. If, A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k. What kind of curve is passing through (0, k)?
a) Parabola
b) Hyperbola
c) Ellipse
d) Circle
Answer: d
Clarification: Equation of the normal at a point P(x, y) is given by
Y – y = -1/(dy/dx)(X – x) ….(1)
Let the point Q at the x-axis be (x1 , 0).
From (1), we get
y(dy/dx) = x1 – x ….(2)
Now, giving that PQ 2 = k 2
Or, x1 – x + y 2 = k 2
=>y(dy/dx) = ± √(k 2 – y 2 ) ….(3)
(3) is the required differential equation for such curves,
Now solving (3) we get,
∫-dy/√(k 2 – y 2 ) = ∫-dx
Or, x 2 + y 2 = k 2 passes through (0, k)
Thus, it is a circle.


2.E: Equations of First Order (Exercises)

Answer the following to the best of your ability. Questions left blank are not counted against you. When you have completed every question that you desire, click the " MARK TEST " button after the last exercise. A new page will appear showing your correct and incorrect responses. If you wish, you may return to the test and attempt to improve your score. If you are stumped, answers to numeric problems can be found by clicking on "Show Solution" to the right of the question.

Do NOT type units into the answer boxes, type only the numeric values. Do NOT use commas or scientific notation when entering large numbers. Answer all non-integer questions to at least 3 significant figures. Correct answers MUST be within ± 1 unit of the third significant figure or they are scored as wrong.


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Solve Simple Differential Equations

y ' = f(x) A set of examples with detailed solutions is presented and a set of exercises is presented after the tutorials. Depending on f(x), these equations may be solved analytically by integration. In what follows C is a constant of integration and can take any constant value.

Exemplo 1: Solve and find a general solution to the differential equation.
y ' = 2x + 1
Solution to Example 1:
Integrate both sides of the equation.
y ' dx = (2x + 1) dx
which gives
y = x 2 + x + C.
As a practice, verify that the solution obtained satisfy the differential equation given above.

Exemplo 2: Solve and find a general solution to the differential equation.
2 y ' = sin(2x)
Solution to Example 2:
Write the differential equation of the form y ' = f(x).
y ' = (1/2) sin(2x)
Integrate both sides
y ' dx = (1/2) sin(2x) dx
Let u = 2x so that du = 2 dx, the right side becomes
y = (1/4) sin(u) du
Which gives .
y = (-1/4) cos(u) = (-1/4) cos (2x)

Exemplo 3: Solve and find a general solution to the differential equation.
y 'e -x + e 2x = 0
Solution to Example 3:
Multiply all terms of the equation by e x and write the differential equation of the form y ' = f(x).
y ' = - e 3x
Integrate both sides of the equation
y ' dx = - e 3x dx
Let u = 3x so that du = 3 dx, write the right side in terms of u
y = (-1/3) e u du
Which gives .
y = (-1/3) e u = (-1/3) e 3x

Solve the following differential equations.
a) 2y ' = 6x
b) y ' cos x = sin(2x)
c) y ' e x = e 3x

Solutions to the above exercises
a) y = (3/2) x 2 + C
b) y = -2 cos x + C
c) y =(1 / 2) e 2x + C


2.E: Equations of First Order (Exercises)

Express [egin x_1' &= 2tx_1 + e^tx_2 x_2' &= 3x_1 - 3t x_2end] with (x_1(0) = -5) and (x_2(0) = 2) as a vector equation with a vector initial condition.

Exercise 2

Consider the system [egin x' &= 2x+y - z y' &= x-3y+5z z' &= 4x -7y +z.end] Write this system as a vector equation.

Exercise 3

Consider the vector equation (<f x>' = egin 4t & 6t^2 2 & t^3 end <f x>+ egin e^t e^<-t>end) . Write this equation as a system of 2 equations.

For Problems 4-7 recast the higher order linear differential equations as a linear system of first order equations. Find the coefficient matrix (ABld(t)) and forcing (fBld(t)) . If the problem is an initial value problem, then be sure to state the initial condition.

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Consider the second order equation (y'' + p(t)y' + q(t)y = g(t)) with initial conditions (y'(0) = 1) and (y(0) = 2) . Let (x_1 = y) and (x_2 = y') , and then express this second order equation as a system of two first order equations. Be sure to include the initial condition for your system.

Exercise 9

Consider the nth order equation (y^ <(n)>+ a_1(t)y^ <(n-1)>+ a_2(t)y^<(n-2)>+ dots + a_(t)y' + a_n(t) y= g(t)) with initial conditions (y^<(i)>(0) = b_) for (i = 0, dots, n-1) . Express this nth order equation as a system of (n) first order equations. Be sure to include the initial condition for your system.

Exercise 10

etiqueta Two hanging pendula of length (ell) and mass (m_1) and (m_2) are coupled by a spring. Let ( heta_1) and ( heta_2) be the angular displacement of each pendulum from its rest position. For small angles, the equations of motion are approximated by the following linear system [egin m_1ell heta_1^ &= - m_1 g heta_1 - kell( heta_1 - heta_2) m_2ell heta_2^ &= - m_2 g heta_2 + kell( heta_1 - heta_2). end] Write this as a first order linear system. Find the corresponding coefficient matrix (ABld) and the forcing (fBld) .

For problems 11-12 determine the largest interval where a unique solution exists for the following initial value problems for first order systems.


Equação

in mathematics, the analytical form of the problem of finding those inputs for which two given functions have equal outputs. The functional inputs are usually called unknowns, and the unknowns for which the functional outputs are equal are called solutions or roots. We say that the solutions satisfy the given equation. For example 3x &ndash 6 = 0 is an equation in one unknown with solution x = 2, and x 2 + y 2 = 25 is an equation in two unknowns, one of whose solutions is x = 3, y = 4. The totality of solutions of an equation depends on the set M of values that may be assigned to the unknowns. An equation may have no solution in M, or it may have some, or even infinitely many, solutions in M. For example, the equation x 4 &ndash 4 = 0 has no rational solutions, two real solutions , and four complex solutions . The equation sin x = 0 has infinitely many real solutions xk = k&pi (k = 0, ±1, ±2, . . .). If an equation is satisfied by every number in M, then we say that it is an identity over M. Por exemplo, is an identity over the non-negative reals but not over the reals.

When we are looking for values of the unknowns that satisfy a set of equations, then we call this set a system of equations, and the numbers in question, solutions of the system. For example, the equations x + 2y = 5, 2x + y &ndash z = 1 are a system or two equations in three unknowns. One solution of this system is x = 1, y = 2, z = 3.

Two systems (two equations) are said to be equivalent if every solution of one system (one equation) is a solution of the other system (the other equation) and conversely. Here we require that both systems (equations) be considered over the same input domain. For example, the equations x &ndash 4 = 0 and 2x &ndash 8 = 0 are equivalent, since their common solution is x = 4. Every system of equations is equivalent to a system of the form fk (x1, x2, . xn) = 0, k = 1, 2, . . . . When finding the solutions of an equation, we usually replace it with an equivalent equation. In some cases it is necessary to replace the given equation with an equation that has more solutions than the given equation. A solution of the new equation that is not a solution of the given equation is called extraneous (seeEXTRANEOUS ROOT). For example, by squaring the equation , we obtain the equation x &ndash 3 = 4 whose solution x = 7 is extraneous for the original equation. Thus, if in the process of solving an equation we resort to steps&mdashsuch as squaring&mdashthat may introduce extraneous roots, then the solutions of the transformed equation must be verified by substitution in the original equation.

We know most about equations for which the functions fk are polynomials in variables x1, x2. xn, that is, algebraic equations. For example, an algebraic equation in one unknown has the form

(*) uma0x n + uma1x n&ndash1 + . + uman = 0 (uma0 &ne 0)

The number n is called the degree of the equation. Solutions of equations of degree 1 and 2 were known in antiquity. The problem of solving an algebraic equation of degree n was one of the most important problems of the 16th and 17th centuries. During that time, mathematicians developed methods for solving equations of degrees 3 and 4 and obtained formulas for the roots of such equations (seeALGEBRA e CARDANO&rsquoS FORMULA). No such formulas exist for equations of degree n &ge 5, since, in general, such equations cannot be solved in radicals. This fact was proved by N. Abel in 1824. About 1830, consideration of the problem of the solvability of algebraic equations in radicals led E. Galois to a general theory of algebraic equations.

Every algebraic equation has at least one real or complex solution. This assertion is the fundamental theorem of algebra, first proved by K. Gauss in 1799. If &alpha is a solution of the equation (*), then the polynomial uma0x n + uma1x n&ndash1 + . + uman is divisible by x &ndash &alpha. If it is divisible by (x &ndash &alpha) k but not by (x &ndash &alpha) K + 1 , then we say that a is a root of multiplicity k. Counting multiplicities, the number of solutions of the equation (*) is n.

Se a função f(x) is transcendental, then the equation f(x) = 0 is called a transcendental equation (as an example, seeKEPLER&rsquoS EQUATION). Depending on the form of f(x), we distinguish trigonometric, logarithmic, and exponential equations. We also consider equations with irrationalities, that is, expressions under radicals. The practical solution of equations involves various approximation methods.

The simplest system of equations is a system of linear equations in which the fk are polynomials of degree 1 in x1, x2, . xn (seeLINEAR EQUATION).

In general, the solution of a system of (not necessarily linear) equations reduces to the solution of a single equation by means of the method of elimination of unknowns (see alsoRESULTANT).

In analytic geometry we interpret an equation in two unknowns as a plane curve consisting of all points whose coordinates satisfy the given equation. We interpret an equation in three unknowns as a surface in 3-space. In this interpretation, the solution of a system of equations reduces to the problem of finding the points of intersection of curves, surfaces, and so on. Equations in more than three unknowns must be interpreted as sets of points in n-dimensional spaces.

In number theory we consider indeterminate equations, that is, equations in several unknowns for which we wish to find solutions that are integers or rational numbers (seeDIOPHANTINE EQUATIONS). For example, the integer solutions of the equation x 2 + y 2 = z 2 are of the form x = m 2 &ndash n 2 y = 2mn z = m 2 + n 2 , where m e n are integers.

Let F and &Phi be mappings of a set UMA into a set B. From the most general point of view, an equation is the statement of the problem of finding elements uma in UMA de tal modo que F(a) = &Phi(a). Se UMA e B are sets of numbers, then we obtain an equation of the type considered above. Se UMA e B are sets in multidimensional spaces, then we obtain systems of equations. Finally, if UMA e B are sets of functions, then, depending on the nature of the mappings, we can also obtain differential, integral and other forms of equations (seeDIFFERENTIAL EQUATIONS e INTEGRAL EQUATIONS). In addition to the problem of finding solutions of an equation, we consider problems of existence and uniqueness of a solution as well as the problem of the continuous dependence of a solution on various data.

The term &ldquoequation&rdquo is also used in other natural sciences in a sense different from&rsquo that above. Relevant examples are the time equation in astronomy, the equation of state in physics, chemical equations, Maxwell&rsquos equations in electrodynamics, and the Boltzmann kinetic equation in the theory of gases.


First order differential equations Calculator

Exemplo

Solved Problems

Difficult Problems

Solved example of first order differential equations

Take $frac<5><4>$ out of the fraction

Group the terms of the differential equation. Move the terms of the $y$ variable to the left side, and the terms of the $x$ variable to the right side

Integrate both sides of the differential equation, the left side with respect to $y$, and the right side with respect to $x$

Applying the power rule for integration, $displaystyleint x^n dx=frac<>>$, where $n$ represents a number or constant function, in this case $n=1$

Solve the integral $int ydy$ and replace the result in the differential equation

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

Apply the power rule for integration, $displaystyleint x^n dx=frac<>>$, where $n$ represents a number or constant function, such as $2$

Simplify the fraction $frac<5><4>left(frac><3> ight)$

Solve the integral $intfrac<5><4>x^2dx$ and replace the result in the differential equation

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

Eliminate the $frac<1><2>$ from the left, multiplying both sides of the equation by $

Removing the variable's exponent

As in the equation we have the sign $pm$, this produces two identical equations that differ in the sign of the term $sqrt<2left(frac<5><12>x^<3>+C_0 ight)>$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign


I could try to add the bushes and trees, to get 19 bushes and 6 trees, but this wouldn't get me anywhere, because I don't have subtotals for the bushes and trees. So I'll pick variables:

With these variables, I can set up a system of equations each equation will represent one of the transactions they've given me:

Multiplying the second row by &ndash2 , I get:

Adding down the t -terms cancel out, leaving me with B = 23 . Back-solving, I get that t = 47 . Of course, the exercise didn't ask for the values of the two variables. Translating back into English, my solution is:

You probably remember the "distance" word problems where you had a boat going with the current and then against the current, or a plane going with the wind (that is, having a tailwind) and then against the wind (that is, having a headwind). Once you learn how to solve systems of equations, you'll see more of these sorts of exercises.

A passenger jet took three hours to fly 1800 miles in the direction of the jetstream. The return trip against the jetstream took four hours. What was the jet's speed in still air and the jetstream's speed?

When they ask me about the speed "in still air" (for planes) or "in still water" (for boats), they are referring to the speedometer reading they are referring only to the powered input, irrespective of outside influences.

On some very windy day, you can watch birds flapping frantically in the air, trying to cross the street, say, from the tree in your front yard. No matter how hard they flapped, they made little or no forward progress sometimes, a bird will even appear to fly backwards! Did this mean that the bird wasn't actually flapping? No it meant that the bird's attempted speed (how fast the flapping would have moved the bird on a windless day) was not fast enough to usefully counteract the wind hitting it in the face. The bird's "speed in still air", less the wind's speed in the opposite direction, was close to zero, or even negative.

The same concept applies to machinery. If a boat's motor is chugging away at 10 miles an hour (according to the speedometer), but the boat is facing a water current of 15 miles an hour in the opposite direction, then the boat will end up going backwards at five miles an hour. In other words, the speedometer reading is not always the actual speed.

Returning to the exercise:

I'll pick variables and set up a system. In this case, I'll use:

When the plane is going "with" the wind, the plane's powered speed and the windspeed will add together when the plane is going "against" the wind, the windspeed will be subtracted from the plane's speedometer reading (that is, from the engines' actual output).

In each case, the "distance" equation will be "(the combined speed) times (the time spent at that speed) equals (the total distance travelled)":

with the jetstream: (p + C)(3) = 1800

against the jetstream: (p &ndash C)(4) = 1800

Rather than multiply through, I notice that, if I divide off the 3 and the 4, I'll have a system that's already set for solving by addition:

Then, by adding down, I get:

Back-solving, I see that the windspeed C must be 75 mph.

Another topic you might see (if not now, then later in calculus) is decomposing rational expressions using partial fractions.

Find the partial fraction decomposition of the following:

The denominator of the polynomial fraction they've given me factors as:

These factors will be the denominators in the partial-fraction decomposition. That is, I'll be looking for the values of A , B , and C which will complete the following:

The above expression is meant to be equal to the original fraction they gave me. Setting them equal and then multiplying both sides by the common denominator, I get:

The standard way of solving this big messy equation is the process of "comparing coefficients". Two polynomials are equal only if the coefficients of their terms are equal. This is why I grouped my terms the way I did in the last line above I've grouped everything that was multiplied by x 2 , everything that was multiplied by x , and everything that was multiplied by 1 (that is, everything that was just a constant, with no variable part).

On the left-hand side, I've got 5x + 7 , which has no term with an x 2 , so I'll need to think of " 5x + 7 " as " 0x 2 + 5x + 7 ". This will allow me to create new equations, based on the fact that the coefficients on either side of the "equals" sign have to be the same. This gives me:

Solving this system, I get A = &ndash1 , B = &ndash1 , and C = 2 . Then the partial fraction decomposition is:

Whatever you do, don't panic when you face a systems-of-equations word problem. If you take them step-by-step, they're usually pretty do-able. That said, it would probably be to your benefit if you did extra practice problems, just to help you get in the swing of things. With any luck, your tests will then go a little faster.


4.5 First-order Linear Equations

Earlier, we studied an application of a first-order differential equation that involved solving for the velocity of an object. In particular, if a ball is thrown upward with an initial velocity of v 0 v 0 ft/s, then an initial-value problem that describes the velocity of the ball after t t seconds is given by

This model assumes that the only force acting on the ball is gravity. Now we add to the problem by allowing for the possibility of air resistance acting on the ball.

Air resistance always acts in the direction opposite to motion. Therefore if an object is rising, air resistance acts in a downward direction. If the object is falling, air resistance acts in an upward direction (Figure 4.24). There is no exact relationship between the velocity of an object and the air resistance acting on it. For very small objects, air resistance is proportional to velocity that is, the force due to air resistance is numerically equal to some constant k k times v . v . For larger (e.g., baseball-sized) objects, depending on the shape, air resistance can be approximately proportional to the square of the velocity. In fact, air resistance may be proportional to v 1.5 , v 1.5 , or v 0.9 , v 0.9 , or some other power of v . v .

The differential equation in this initial-value problem is an example of a first-order linear differential equation. (Recall that a differential equation is first-order if the highest-order derivative that appears in the equation is 1. ) 1. ) In this section, we study first-order linear equations and examine a method for finding a general solution to these types of equations, as well as solving initial-value problems involving them.

Definition

A first-order differential equation is linear if it can be written in the form

Examples of first-order nonlinear differential equations include

Standard Form

Consider the differential equation

Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of y ′ y ′ be equal to 1 . 1 . To make this happen, we divide both sides by 3 x 2 − 4 . 3 x 2 − 4 .

This is called the standard form of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation. Returning to Equation 4.14, we can divide both sides of the equation by a ( x ) . a ( x ) . This leads to the equation

We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.

Example 4.15

Writing First-Order Linear Equations in Standard Form

Put each of the following first-order linear differential equations into standard form. Identify p ( x ) p ( x ) and q ( x ) q ( x ) for each equation.


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