Question

The solution of the equation $\frac{dy}{dx}+x(x+y)=$ $x^3(x+y{)}^3-1$ is

• A. $(x+y{)}^{-3}=c{e}^{x^2}+x^2+1$
• B. $(x+y{)}^{-2}=c{e}^{x2}-x^2+1$
• C. $(x+y{)}^{-2}=c{e}^{x2}+x^2+1$
• D. None of these

Answer 1

Answer: (C)

The given equation can be written as
$\left(\frac{dy}{dx}+1\right)+x(x+y)=x^3(x+y{)}^3$
$\Rightarrow \frac{d(x+y)}{dx}+x(x+y)=x^3(x+y{)}^3$
$\Rightarrow (x+y{)}^3\frac{d(x+y)}{dx}+x(x+y{)}^{-2}=x^3$
Let $(x+y{)}^{-2}=z$ so that
$-2(x+y{)}^{-3}\frac{d(x+y)}{dx}=\frac{dz}{dx}$
The given equation reduces to
$\frac{-1}{2}\frac{dz}{dx}+xz=x^3$
i.e., $\frac{dz}{dx}-2xz=-2x^3$
I.F. $=e^{\int -2xdx}=e^{-x2}$
$\therefore$ The solution is,
$z\cdot e^{-x^2}=\int -2x^3\cdot e^{-x^2}dx=\left(x^2+1\right)e^{-x^2}+c$
or, $\frac{1}{(x+y{)}^2}=c{e}^{x^2}+x^2+1$