Question

Solution of the differential equation $\frac{dy}{dx}+x(x+y)=x^3(x+y{)}^3-1$ is:
(where $C$ is integration constant)

• A. $\frac{1}{(x+y)}=C{e}^x+x^2-1$
• B. $\frac{1}{(x+y)}=C{e}^{x^2}+x^3+1$
• C. $\frac{1}{(x+y)}=C{e}^{x^2}+x^2+1$
• D. $\frac{1}{(x+y{)}^2}=C{e}^{x^2}+x^2+1$

Answer 1

The correct option is D $\frac{1}{(x+y{)}^2}=C{e}^{x^2}+x^2+1$

The given Equation can be written as $(\frac{dy}{dx}+1)+x(x+y)=x^3(x+y{)}^3$
$\Rightarrow \frac{d(x+y)}{dx}+x(x+y)=x^3(x+y{)}^3$
Dividing both sides by $(x+y{)}^3$ we have:
$\Rightarrow (x+y{)}^{-3}\cdot \frac{d(x+y)}{dx}+x(x+y{)}^{-2}=x^3$
Let $(x+y{)}^{-2}=z$ so that $-2(x+y{)}^{-3}\frac{d(x+y)}{dx}=\frac{dz}{dx}$
The given equation now reduces to
$-\frac{1}{2}\frac{dz}{dx}+xz=x^3$
$\Rightarrow \frac{dz}{dx}-2xz=-2x^3$
On comparing with $\frac{dz}{dx}+Pz=Q,$
We get $P=-2x$ and $Q=-2x^3$
I.F. $=e^{\int -2xdx}=e^{-x^2}$

$\therefore$ The solution is:
$z\cdot e^{-x^2}=\int -2x^3\cdot e^{-x^2}dx$
Substituting $-x^2=t$ and later integrating with By parts, we have:
$\Rightarrow z\cdot e^{-x^2}=(x^2+1)e^{-x^2}+C$
$\Rightarrow \frac{1}{(x+y{)}^2}=C{e}^{x^2}+x^2+1$