Question

The solution of the differential equation $\frac{dy}{dx}+x\left(2x+y\right)=x^3{\left(2x+y\right)}^3-2$ is ( $C$ being an arbitrary constant)

• A. $\frac{1}{2x+xy}=x^2+1+C{e}^x$
• B. $\frac{1}{{\left(2x+y\right)}^2}=x^2+1+C{e}^{x^2}$
• C. $\frac{1}{2x+y}=x+1+C{e}^{-x^2}$
• D. $\frac{1}{{\left(2x+y\right)}^2}=x^2+1+C$

Answer 1

Answer: (B)

Let, $2x+y=t\Rightarrow \frac{dy}{dx}+2=\frac{dt}{dx}$
$\frac{dt}{dx}+xt=x^3{t}^3\Rightarrow \frac{1}{t^3}\frac{dt}{dx}+\frac{1}{t^2}x=x^3$
Let, $\frac{1}{t^2}=u\Rightarrow \frac{-2}{t^3}\frac{dt}{dx}=\frac{du}{dx}$
$\frac{du}{dx}+\left(-2x\right)u=-2x^3$
$I.F.=e^{-\int 2xdx}=e^{-x^2}\Rightarrow u.e^{-x^2}=\int e^{-x^2}\left(-2x^3\right)dx$
$\frac{e^{-x^2}}{{\left(2x+y\right)}^2}=-2\int e^{-x^2}.x^{3}dx$
$\frac{e^{-x^2}}{{\left(2x+y\right)}^2}=\int e^{-x^2}.x^2\left(-2x\right)dx$
Let, $-x^2=v$
$-2xdx=dv\Rightarrow \frac{e^{-x^2}}{(2x+y{)}^2}=-\int e^{v}vdv$
$\frac{e^{-x^2}}{(2x+y{)}^2}+v\cdot e^v-e^v=C\Rightarrow \frac{e^{-x^2}}{(2x+y{)}^2}-x^2{e}^{-x^2}-e^{-x^2}=C$
$\frac{1}{(2x+y{)}^2}=\left(x^2+1\right)+C{e}^{x^2}$