Question

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

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

Answer 1

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

$\frac{dy}{dx}+x(x+y)=x^3(x+y{)}^3-1\Rightarrow (\frac{dy}{dx}+1)+x(x+y)=x^3(x+y{)}^3\Rightarrow (x+y{)}^{-3}\frac{d(y+x)}{dx}+x(x+y{)}^{-2}=x^3$

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

$\Rightarrow -\frac{1}{2}\frac{dz}{dx}+xz=x^3\Rightarrow \frac{dz}{dx}-2xz=-2x^3$

So, integrating factor is,
$I.F.=e^{\int -2xdx}=e^{-x^2}$
Now,
$z\cdot e^{-x^2}=\int -2x^3{e}^{-x^2}dx+c$
Let $I=\int 2x^3{e}^{-x^2}dx$
Put $x^2=t\Rightarrow 2xdx=dt$
$I=\int t{e}^{-t}dt=-t{e}^{-t}-e^{-t}$
So,
$\Rightarrow z\cdot e^{-x^2}=e^{-t}(1+t)+C\Rightarrow \frac{1}{(x+y{)}^2}=C{e}^{x^2}+1+x^2$