Question

Solução particular e complementar da EDO : 2 . dy/dx + 4 = 6; y(0) = 3/2 (o 2 multiplica dy/dx)

Answer 1

Olá

Dada a EDO 1ª Ordem

$\displaystyle \mathsf{2 \cdot\frac{dy}{dx}+4y=6 \qquad\qquad y(0)= \frac{3}{2} }\\\\\\\mathsf{2y'+4y=6}$

O y' não pode estar acompanhado de outro termo, então, vamos dividir toda a EDO por 2 para eliminar o 2 que está junto ao y'

$\displaystyle \mathsf{ \frac{2y'}{2} + \frac{4y}{2} = \frac{6}{2} }\\\\\\\mathsf{y'+2y=3}$

Calculando o fator integrante

$\displaystyle \mathsf{I_{(x)}=e^{\int 2dx}~=~\boxed{e^{2x}}}$

Multiplicando a EDO pelo fator integrante

$\displaystyle \mathsf{e^{2x}y'+e^{2x}2y=e^{2x}3}\\\\\\\underbrace{(\mathsf{e^{2x}y'+e^{2x}2y})}_{(y\cdot e^{2x})'}$

Integrando dos dois lados

$\displaystyle \mathsf{\int (y\cdot e^{2x})'dx ~=~\int 3e^{2x}dx}\\\\\\\mathsf{\diagup\!\!\!\!\!\!\!\int (y\cdot e^{2x})^{\diagup\!\!\!\!}'dx~=~3\int e^{2x}dx}\\\\\\\mathsf{ye^{2x}~= ~ \frac{3e^{2x}}{2}+C }$


Isolando o y

$\displaystyle \mathsf{y~= ~ \frac{3e^{2x}}{2e^{2x}}+ \frac{C}{e^{2x}} }\\\\\\\boxed{\mathsf{y= \frac{3}{2} + \frac{C}{e^{2x}} }}$


Dados do valor inicial

y(0) = 3/2
x = 0    ;     y = 3/2

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

$\displaystyle \mathsf{y= \frac{3}{2} + \frac{0}{e^{2x}} }\\\\\\\boxed{\mathsf{y= \frac{3}{2} }}$