Question

The solution of the differential equation, $e^x\left(x+1\right)dx+\left(y{e}^y+x{e}^x\right)dy=0$ with initial condition $f\left(0\right)=0,$ is-

• A. $x{e}^x+2y^2{e}^y=0$
• B. $2x{e}^x+y^2{e}^y=0$
• C. $x{e}^x-2y^2{e}^y=0$
• D. $2x{e}^x-y^2{e}^y=0$

Answer 1

The correct option is A $x{e}^x+2y^2{e}^y=0$

sol $\Rightarrow$ To solve differentiated equation

we write

$Mdx+Ndx=0$

Where $M=l^x(x+1)$

$N=y{e}^y+x{e}^x$

So $\frac{\partial M}{\partial y}=0$ and $\frac{\partial N}{\partial x}=o+[x{e}^x+e^x]=e^x[x+1]$

As $\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$

$\Rightarrow \frac{1}{M}[\frac{\partial N}{\partial M}-\frac{\partial M}{\partial y}]=\frac{1}{e^x(x+1)}\left[e^x\right(x+1)-0]=1=g\left(y\right)$

$\Rightarrow 1.F.=e^{\int g\left(y\right)dx}=e^{\int 1dy}=e^y$

$e^x(x+1)dx+(y{e}^y+x{e}^x)dy=0$

$\Rightarrow e^x{e}^y(x+1)dy+e^y(y{e}^y+x{e}^x)dy=0$

$f(x,y)=\underset{y=\cos 5t}{\int }mdx+\int N(\ne x)dy=c$

$=\int e^x{e}^y(x+1)dx+\int e^y.y{e}^{y}dy=c$

$=e^y\int e^x(x+1)dx+\int e^{2y}ydy=e$

$=x{e}^x.e^y+e^y(y-1)=c$

So at $x=o$ and $y=0$

$=0.11+1\left(0.1\right)=c\Rightarrow c=-1$

$f(x,y)=x{e}^{x+y}+e^y(y-1)-c=0$

$=x{e}^{x+y}+e^y(y-1)+1=0$