Question

Solution of the differential equation, $\left(x{y}^2-e^{1/x^3}\right)dx=x^{2}ydy$ is

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

Answer 1

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

Given,

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

$x^{2}y\frac{dy}{dx}-x{y}^2=-e^{\frac{1}{x^3}}$$-\left(1\right)$

let $y^2=t$

$\therefore 2ydy=dt$

$\left(1\right)\to \frac{x^{2}dt}{2dx}-xt=-e^{-\frac{1}{x^3}}$

$\frac{dt}{dx}-\frac{2t}{x}=-\frac{2}{x^2}{e}^{-\frac{1}{x^3}}$

P=$\frac{-2}{x}$,Q=$-\frac{2}{x^3}{e}^{-\frac{1}{x^3}}$

I.f.=$e^{\int p.dx}$

=$e^{\int \frac{-2}{x}dx}$

=$e^{-2logx}$

=$elog{x}^{-2}$

=$x^{-2}$

Com-plete solution

$t\times x^{-2}=\int x^{-2}\times -\frac{2}{x^2}{e}^{\frac{-1}{x^3}}dx+c$

$t\times x^{-2}=-\int \frac{2}{x^4}{e}^{\frac{-1}{x^3}}dx+c$

Let$\frac{-1}{x^3}=u$

$\to \frac{3}{x^4}dx=du$

$t\times x^{-2}=-\int \frac{2}{3}{e}^{u}du+c$

$t\times x^{-2}=-\frac{2}{3}{e}^u+c$

Substitute the value of t and u

$\therefore y^2\times x^{-2}=-\frac{2}{3}{e}^{\frac{-1}{x^3}}+c$

$\frac{y^2}{2x^2}+\frac{1}{3}{e}^{\frac{-1}{x^3}}=c$