Question

Solve the differential equations: $\frac{dy}{dx}+yx=y^2.e^{x^2/2}\sin x$ or $y^2{e}^{x^2/2}\log x$

• A. $y{e}^{x^2/2}(c+\cos x)+1=0$ or $y{e}^{x^2/2}(x\log x-x+c)+1=0.$
• B. $y{e}^{x^2/2}(c-\cos x)+1=0$ or $y{e}^{x^2/2}(x\log x-x+c)+1=0.$
• C. $y{e}^{x^2/2}(c-\cos x)+1=0$ or $y{e}^{x^2/2}(x\log x+x+c)+1=0.$
• D. $y{e}^{x^2/2}(c-\cos x)+1=0$ or $y{e}^{x^2/2}(-x\log x-x+c)+1=0.$

Answer 1

Answer: (B)

$y{e}^{x^2/2}(c-\cos x)+1=0$ or $y{e}^{x^2/2}(x\log x-x+c)+1=0.$

$\frac{dy}{dx}+yx=y^2.e^{\frac{x^2}{2}}.\sin x\Rightarrow \frac{1}{y^2}\frac{dy}{dx}+\frac{yx}{y^2}=e^{\frac{x^2}{2}}\sin x$
$\Rightarrow \frac{1}{y^2}\frac{dy}{dx}+\frac{1}{y}x=e^{\frac{x^2}{2}}\sin x$
Put $-\frac{1}{y}=v\Rightarrow \frac{1}{y^2}dy=dv$
$\therefore \frac{dv}{dx}-vx=e^{\frac{x^2}{2}}\sin x$ ...(1)
Here, $P=-x\Rightarrow \int Pdx=-\int xdx=\frac{-x^2}{2}$
$\therefore I.F.=e^{-\frac{x^2}{2}}$
Multiplying $1$ by $I.F.$, we get
$e^{-\frac{x^2}{2}}\frac{dv}{dx}-v{e}^{-\frac{x^2}{2}}=\sin x$
Integrating both sides we get,
$e^{-\frac{x^2}{2}}v=\cos x+C$ $\Rightarrow v=e^{\frac{x^2}{2}}(\cos x+C)\Rightarrow y{e}^{\frac{x^2}{2}}(\cos x+C)=-1$
Now similarly for
$\frac{dy}{dx}=y^2{e}^{\frac{x^2}{2}}\log x$
$\Rightarrow -\frac{1}{y}.e^{-\frac{x^2}{2}}=x\log x-x+C\Rightarrow y{e}^{\frac{x^2}{2}}(x\log x-x+C)+1=0$