Question

The solution of $\frac{dy}{dx}=e^{3x+y}$ given $y=0$ when $x=0$ is:

• A. $e^{3x}+3e^{-y}=4$
• B. $e^{-y}=e^{3x}+4$
• C. $3e^{-y}=e^{3x}+12$
• D. $y=\frac{e^{3x+y}}{3}$

Answer 1

The correct option is A $e^{3x}+3e^{-y}=4$

$\int e{}^{-y}dy=\int e{}^{3x}dx+c$
$\Rightarrow \frac{e{}^{-y}}{-1}=\frac{e{}^{3x}}{3}+c$
$\Rightarrow 3e{}^{-y}+e{}^{3x}=c$
Put $(x,y)=(0,0)$
$\Rightarrow 3+1=4=c$
$\Rightarrow 3e{}^{-y}+e{}^{3x}=4$