Question

EQUAÇÃO DIFERENCIAL Y'=1/e^3X COM VALOR INICIAL Y(0)=0.

Answer 1

$y' = \frac{1}{e^{3x}} \\\\\text{entao}\\\\ \frac{dy}{dx} = \frac{1}{e^{3x}}\\\\dy = \frac{1}{e^{3x}} dx\\\\ \boxed{\int dy= \int \frac{1}{e^{3x}} dx}$

resolvendo 
$\int 1 dy = y$

a segunda integral
$\int \frac{1}{e^{3x}} dx = \int e^{-3x} dx}$

fazendo substituição 
$u= -3x\\\\ \frac{du}{dx}= -3 \to \frac{-1}{3}du = dx$

substituindo os valores
$\int e^u * \frac{-1}{3} du\\\\\\=- \frac{1}{3}\int e^u du \\\\= \frac{-1}{3}e^u +C\\\\=\boxed{\boxed{ \frac{-e^{-3x}}{3}+C }}$

temos
$y = \frac{-e^{-3x}}{3}+C }$

calculando y(0) = 0  -> quando x=0 , y=0

$0 = \frac{-e^{-3*0}}{3} +C\\\\ 0 = \frac{-e^0}{3}+C \\\\0= \frac{-1}{3}+C\\\\\boxed{\boxed{C= \frac{1}{3} }}$

então a função procurada é 

$y= \frac{-e^{-3x}}{3}+ \frac{1}{3} \\\\\boxed{\boxed{y= \frac{1-e^{-3x}}{3} }}$