Question

For differential equation $2\frac{dy}{dx}-\frac{2y}{x+1}=\frac{x^3}{y^2}.$ The solution is $y^3{(1+x)}^2=\frac{1}{A}{x}^8+\frac{2}{5}{x}^5+\frac{1}{B}{x}^4+c.$ Find A+B

Answer 1

$2\frac{dy}{dx}-\frac{2y}{x+1}=\frac{x^3}{y^2}\Rightarrow 2y^2\frac{dy}{dx}-\frac{2y^3}{x+1}=x^3$
Put $y^3=v\Rightarrow 3y^{2}dy=dv$
$\therefore 6\frac{dv}{dx}-\frac{2v}{x+1}=x^3$ ...(1)
Here, $P=-\frac{2}{x+1}\Rightarrow \int Pdx=-\int \frac{2}{x+1}dx=-2\log (x+1)=\log {(x+1)}^{-2}$
$\therefore I.F.=e\log {(x+1)}^{-2}=\frac{1}{(x+1{)}^2}$
Multiplying $1$ by $I.F.$, we get
$\frac{6}{(x+1{)}^2}\frac{dv}{dx}-\frac{2v}{(x+1{)}^3}=\frac{x^3}{(x+1{)}^2}$
Integrating both the sides
$v\frac{1}{(x+1{)}^3}=\int \frac{x^3}{(x+1{)}^2}dx+C$
$\Rightarrow y^3{(1+x)}^2=\frac{1}{6}{x}^8+\frac{2}{5}{x}^5+\frac{1}{4}{x}^4+C$
$\therefore A+B=10$