Question

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

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

Answer 1

Answer: (B)

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