Question

The solution of the differential equation, $e^x(x+1)dx+(y{e}^y+x{e}^x)dy=0$ 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 B $2x{e}^x+y^2{e}^y=0$

$(e^{y}y+e^{x}x)\frac{dy}{dx}+e^x(x+1)=0$ ...(1)
Let $R(x,y)=e^x(x+1)$ and $S(x,y)=e^{y}y+e^{x}x$
This is not an exact equation, because $\frac{dR(x,y)}{dy}=0\ne e^{x}x+e^x=\frac{dS(x,y)}{dx}$
Find an integrating factor $\mu$ such that $\mu R(x,y)+\mu \frac{dy}{dx}S(x,y)=0$ is exact
This means $\frac{d}{dy}\left(\mu R(x,y)\right)=\frac{d}{dx}\left(\mu S(x,y)\right)$
$\Rightarrow \frac{d\mu }{dy}{e}^x(x+1)=(e^{x}x+e^x)\mu \Rightarrow \frac{\frac{d\mu }{dy}}{\mu }=1\Rightarrow \log \mu =y\Rightarrow \mu =e^y$
Multiply both side of (1) by $\mu$
$e^y(e^{y}y+e^{x}x)\frac{dy}{dx}+e^{x+y}(x+1)=0$
Let $P(x,y)=e^{x+y}(x+1)$ and $Q(x,y)=e^y(e^{y}y+e^{x}x)$
This is an exact equation, because $\frac{dP(x,y)}{dy}=e^{x+y}(x+1)=\frac{dQ(x,y)}{dx}$
Define $f(x,y)$ such that $\frac{df(x,y)}{dx}=P(x,y)$ and $\frac{df(x,y)}{dy}=Q(x,y)$
Then the solution will be given by $f(x,y)=c$
Integrate $\frac{df(x,y)}{dx}$ w.r.t.x
$f(x,y)=\int e^{x+y}(x+1)dx=e^{x+y}x+g\left(y\right)$
Differentiating $f(x,y)$ w.r.t y
$\frac{df(x,y)}{dy}=\frac{d}{dy}(e^{x+y}x+g\left(y\right))=e^{x+y}x+\frac{dg\left(y\right)}{dy}$
Substituting into $\frac{df(x,y)}{dy}=Q(x,y)$
$e^{x+y}x+\frac{dg\left(y\right)}{dy}=e^y(e^{y}y+e^{x}x)\Rightarrow \frac{dg\left(y\right)}{dy}=e^{2y}y\Rightarrow g\left(y\right)=\int e^{2y}ydy=e^{2y}(\frac{y}{2}-\frac{1}{4})$
As $f\left(0\right)=0$
hence $2x{e}^x+y^2{e}^y=0$