Question

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

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

Answer 1

Answer: (B)

$\frac{1}{(2x+y{)}^2}=x^2+1+C{e}^{x^2}$

Substitute $2x+y=t\Rightarrow \frac{dy}{dx}+2=\frac{dt}{dx}$

$\frac{dt}{dx}+xt=x^3{t}^3\Rightarrow \frac{1}{t^3}\frac{dt}{dx}+\frac{1}{t^2}x=x^3$

Thus given equation reduced to,

$\frac{1}{t^2}=u\Rightarrow \frac{-2}{t^3}\frac{dt}{dx}=\frac{du}{dx}$

$\frac{du}{dx}+(-2x)u=-2x^3$

$I.F.=e^{-\int 2xdx}=e^{-x^2}\Rightarrow u.e^{-x^2}=\int e^{-x^2}(-2x^3)dx$

$\frac{e^{-x^2}}{(2x+y{)}^2}=-2\int e^{-x^2}.x^{3}dx$

$\frac{e^{-x^2}}{(2x+y{)}^2}=\int e^{-x^2}.x^2(-2x)dx\Rightarrow -x^2=v$

$-2xdx=dv\Rightarrow \frac{e^{-x^2}}{(2x+y{)}^2}=-\int e^{v}vdv$

$\frac{e^{-x^2}}{(2x+y{)}^2}+v.e^v-e^v=C\Rightarrow \frac{e^{-x^2}}{(2x+y{)}^2}-x^2{e}^{-x^2}-e^{-x^2}=C$

$\frac{1}{(2x+y{)}^2}=(x^2+1)+C{e}^{x^2}$