Question

Help! Solução particular da equação diferencial dy/dx + 4y = 12; y(0) =2

Answer 1

Olá

EDO 1ª Ordem
$\mathsf{y'+p(x)y=q(x)}$

Dada a EDO 1ª Ordem

$\displaystyle \mathsf{ \frac{dy}{dx} +4y=12\qquad\qquad y(0)=2}\\\\\\\mathsf{y'+4y=12}$

Calculando o fator integrante

$\mathsf{I_{(x)}=e^{\int p(x)dx}}$

$\displaystyle \mathsf{I_{(x)}=e^{\int 4dx}~=~\boxed{e^{4x}}}$


Multiplicando a EDO pelo fator integrante

$\displaystyle \mathsf{e^{4x}y'+e^{4x}4y=e^{4x}12}$

O que temos no primeiro termo, nada mais é que o produto da derivada entre y e o fator integrante.
Então

$\mathsf{\underbrace{e^{4x}y'+e^{4x}4y}_{(e^{4x}\cdot y)'}=e^{4x}12}$

Integrando dos dois lados

$\displaystyle \mathsf{\int (e^{4x}\cdot y){'}~=~\int 12e^{4x}}\\\\\\ \mathsf{\diagup\!\!\!\!\!\!\!\int (e^{4x}\cdot y){^\diagup\!\!\!'}~=~12\int e^{4x}}\quad\Longrightarrow\quad\boxed{\text{Dica:}\mathsf{\int e^{\alpha x }dx~=~} \frac{e^{\alpha x}}{\alpha}+C\qquad \alpha \in \Re}\\\\\\\mathsf{e^{4x}y=12\cdot \frac{e^{4x}}{4} +C}\\\\\\\text{Simplifica}\\\\\\\mathsf{e^{4x}y=3e^{4x}+C}$

Isolando o y

$\displaystyle\mathsf{y= \frac{3e^{4x}}{e^{4x}} + \frac{C}{e^{4x}} }\\\\\\\mathsf{y=3+ \frac{C}{e^{4x}} }$

Condição do valor inicial

y(0) = 2
x = 0   ;  y = 2

$\displaystyle\mathsf{2=3 + \frac{C}{e^{4\cdot 0}} }\\\\\\\mathsf{2-3= \frac{C}{\underbrace{e^0}}_{=1} }\\\\\\\mathsf{C=-1}$

Resposta

$\displaystyle \boxed{\mathsf{y=3- \frac{1}{e^{4x}} }}\\\\\\~~~ ~~~~\text{ou}\\\\\\\boxed{\mathsf{y=3-e^{-4x}}}$