Question

Solve the differential equation: $(x{y}^2-e^{1/x^3})dx-x^{2}ydy=0,$

• A. $3y^2=2x^2{e}^{1/x^3}+c{x}^2.$
• B. $3y^2=2x^2{e}^{1/x^3}-c{x}^2.$
• C. $3y^2=-2x^2{e}^{1/x^3}+c{x}^2.$
• D. None of these.

Answer 1

The correct option is A $3y^2=2x^2{e}^{1/x^3}+c{x}^2.$

Given, $(x{y}^2-e^{\frac{1}{x^3}})dx-x^{2}ydy=0$

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

$\Rightarrow 3y^2=2x^2{e}^{1/x^3}+c{x}^2$