Question

Solve the differential equation : $x\frac{dy}{dx}=2y+x^4+6x^2$, $x\ne 0$

• A. $y=\frac{x^4}{2}+6x^3\log x+c{x}^2$
• B. $y=\frac{x^4}{2}+6x^2\log x+c{x}^2$
• C. $y=\frac{x^4}{2}-6x^3\log x+c{x}^2$
• D. $y=\frac{y^4}{2}-6x^2\log x+c{x}^2$

Answer 1

The correct option is B $y=\frac{x^4}{2}+6x^2\log x+c{x}^2$

$x\frac{dy}{dx}=2y+x^4+6x^2\Rightarrow \frac{dy}{dx}-\frac{2y}{x}=\frac{x^4+6x^2}{x}$
Let $u=e^{\int -\frac{2}{x}dx}=\frac{1}{x^2}$
Multiply both sides by $u$
$\frac{\frac{dy}{dx}}{x^2}-\frac{2y}{x^3}=\frac{x^4+6x^2}{x^3}$
Substitute $-\frac{2}{x^3}=\frac{d}{dx}\left(\frac{1}{x^2}\right)$
$\frac{\frac{dy}{dx}}{x^2}+\frac{d}{dx}\left(\frac{1}{x^2}\right)y=\frac{x^4+6x^2}{x^3}$
Using $g\frac{df}{dx}+f\frac{dg}{dx}=\frac{d}{dx}\left(fg\right)$
$\frac{d}{dx}\left(\frac{y}{x^2}\right)=\frac{x^4+6x^2}{x^3}$
Integrating both sides w.r.t x, we get
$\int \frac{d}{dx}\left(\frac{y}{x^2}\right)dx=\int \frac{x^4+6x^2}{x^3}dx\Rightarrow \frac{y}{x^2}=\frac{x^2}{2}+6\log x+c\Rightarrow y=\frac{x^4}{2}+6x^2\log x+c{x}^2$