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find the solution of the 1st order linear equations y'+3 y=2x where y(0)=1

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Answer to a math question find the solution of the 1st order linear equations y'+3 y=2x where y(0)=1

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Timmothy
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96 Answers
Para resolver a equação diferencial de 1ª ordem y' + 3y = 2x com a condição inicial y(0) = 1 , vamos usar o método do fator integrante.

Passo 1: Escrever a equação na forma padrão y' + P(x)y = Q(x)
A equação é y' + 3y = 2x , onde P(x) = 3 e Q(x) = 2x .

Passo 2: Calcular o fator integrante
O fator integrante é dado por e^{\int P(x) dx} .
Nesse caso, o fator integrante é e^{\int 3 dx} = e^{3x} .

Passo 3: Multiplicar a equação diferencial pelo fator integrante
Multiplicando a equação y' + 3y = 2x por e^{3x} obtemos:
e^{3x}y' + 3e^{3x}y = 2xe^{3x} .

Passo 4: Integrar ambos os lados da equação resultante
Integrando ambos os lados obtemos:
\int e^{3x}y' dx + \int 3e^{3x}y dx = \int 2xe^{3x} dx .
Integrando, obtemos:
e^{3x}y = \frac{2}{3}xe^{3x} - \frac{2}{9}e^{3x} + C ,
onde C é a constante de integração.

Passo 5: Aplicar a condição inicial
Como y(0) = 1 , podemos substituir na equação acima:
e^{0} \cdot 1 = \frac{2}{3} \cdot 0 \cdot e^{0} - \frac{2}{9} \cdot e^{0} + C .
Isolando C , temos:
1 = -\frac{2}{9} + C ,
C=1+\frac{2}{9}=\frac{11}{9} .

Passo 6: Encontrar a solução da equação diferencial
Substituindo o valor de C na equação obtida, temos:
e^{3x}y=\frac{2}{3}xe^{3x}-\frac{2}{9}e^{3x}+\frac{11}{9} ,
y=\frac{2}{3}x-\frac{2}{9}+\frac{11}{9}e^{-3x} .

\boxed{y=\frac{2}{3}x-\frac{2}{9}+\frac{11}{9}e^{-3x}}

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