Question

Solve the differential equation: $\frac{dy}{dx}+y=x^2{y}^2.$

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

Answer 1

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

$\frac{dy}{dx}+y=x^2{y}^2\Rightarrow \frac{1}{y^2}\frac{dy}{dx}+\frac{1}{y}=x^2$
Put $-\frac{1}{y}=v\Rightarrow \frac{1}{y^2}dy=dv$
$\therefore \frac{dv}{dx}-v=x^2$ ...(1)
Here $P=-1\Rightarrow \int Pdx=-\int dx=-x$
$\therefore I.F.=e^{-x}$
Multiplying (1) by $I.F.$ we get
$e^{-x}\frac{dv}{dx}-e^{-x}v={e^{-x}x}^2$
Integrating both sides we get
$e^{-x}v=\int e^{-x}{x}^{2}dx+c=-e^{-x}{x}^2+\int 2x{e}^{-x}dx+c$

$=-e^{-x}{x}^2-2x{e}^{-x}+\int 2e^{-x}dx+c=-e^{-x}{x}^2-2x{e}^{-x}-2e^{-x}+c$
$\Rightarrow \frac{1}{y}=(x^2+2x+2)-c{e}^x$