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

Answer 1

The correct option is C ${(x+y)}^{-2}=c{e}^{x^2}+x^2+1$

Given, $\frac{dy}{dx}+x(x+y)=x^3{(x+y)}^3-1$

$\Rightarrow (\frac{dy}{dx}+1)+x(x+y)=x^3{(x+y)}^3$

$\Rightarrow \frac{d(x+y)}{dx}+x(x+y)=x^3{(x+y)}^3\Rightarrow \frac{d(x+y)}{{(x+y)}^{3}dx}+x{(x+y)}^{-2}=x^3$

Put ${(x+y)}^{-2}=z\Rightarrow -2{(x+y)}^{-3}\frac{d(x+y)}{dx}=\frac{dz}{dx}$

$\therefore \frac{-1}{2}\frac{dz}{dx}xz=x^3\Rightarrow \frac{dz}{dx}-2xz=-{2x}^3$ ...(1)

Here $P=-2x\Rightarrow \int PdP=\int -2xdx=-x^2$

$\therefore I.F.=e^{{-x}^2}$

Multiplying (1) by $I.F.$ we get

$e^{{-x}^2}\frac{dz}{dx}-e^{{-x}^2}.2xz=-e^{{-x}^2}.{2x}^3$

Integrating both sides, we get

$z.e^{{-x}^2}=\int {-2x}^3.e^{{-x}^2}dx=(x^2+1)e^{{-x}^2}+c$

$\Rightarrow \frac{1}{{(x+y)}^2}={ce}^{x^2}+x^2+1$

${(x+y)}^{-2}=c{e}^{x^2}+x^2+1$