Question

Exercise 5.

Find the solution of

dy/dx= e2x+y 2x+y

that has y = 0 when x = 0

Answer 1

Given equation,

$\dfrac{dy}{dx}=e^{2x+y}$

$\implies \dfrac{dy}{dx}=e^{2x}.e^y$

$\implies e^{-y} dy=e^{2x} dx$

Integrate Both the sides

$\implies \int e^{-y} dy=\int e^{2x} dx$

$\implies -e^{-y}=\dfrac{e^{2x}}{2}+C$

As,y(0)=0,

$\implies -e^{-0}=\dfrac{e^0}{2}+C$

So $C=-1-\dfrac{1}{2}=\dfrac{-3}{2}$

So the solution is

$-e^{-y}=\dfrac{e^{2x}}{2}-\dfrac{3}{2}$

i.e. $e^{-y}=-\dfrac{e^{2x}}{2}+\dfrac{3}{2}$