# What is the general solution of the differential equation y'-1/xy=2x^2\lnx?

## (because Symbolab is addicted to weird $C$ constants...)

Apr 6, 2018

$y = {x}^{3} \ln x - {x}^{3} / 2 + C x$

#### Explanation:

We have:

$y ' - \frac{1}{x} y = 2 {x}^{2} \ln x$

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$\frac{\mathrm{dy}}{\mathrm{dx}} + P \left(x\right) y = Q \left(x\right)$

As the equation is already in this form, then the integrating factor is given by;

$I = {e}^{\int P \left(x\right) \mathrm{dx}}$
$\setminus \setminus = \exp \left(\int \setminus - \frac{1}{x} \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(- \ln x\right)$
$\setminus \setminus = \exp \left(\ln \left(\frac{1}{x}\right)\right)$
$\setminus \setminus = \frac{1}{x}$

And if we multiply the DE by this Integrating Factor, $I$, we will have a perfect product differential;

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

$\therefore \left(\frac{1}{x}\right) \frac{\mathrm{dy}}{\mathrm{dx}} - \left(\frac{1}{x} ^ 2\right) y = 2 x \ln x$

$\therefore \frac{d}{\mathrm{dx}} \left(\frac{y}{x}\right) = 2 x \ln x$

This is now separable, so by "separating the variables" we get:

$\frac{y}{x} = 2 \setminus \int \setminus x \ln x \setminus \mathrm{dx}$

To integrate the RHS integral we can apply integration by Parts:

Let  { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=x, => v,=x^2/2 ) :}

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

We have:

$\int \setminus \left(\ln x\right) \left(x\right) \setminus \mathrm{dx} = \left(\ln x\right) \left({x}^{2} / 2\right) - \int \setminus \left({x}^{2} / 2\right) \left(\frac{1}{x}\right) \setminus \mathrm{dx}$

$\therefore \int \setminus x \ln x \setminus \mathrm{dx} = \frac{{x}^{2} \ln x}{2} - \int \setminus \frac{x}{2} \setminus \mathrm{dx}$
$\text{ } = \frac{{x}^{2} \ln x}{2} - {x}^{2} / 4$

With this result, we have:

$\frac{y}{x} = 2 \left\{\frac{{x}^{2} \ln x}{2} - {x}^{2} / 4\right\} + C$

$y = 2 x \left\{\frac{{x}^{2} \ln x}{2} - {x}^{2} / 4\right\} + C x$
$\setminus \setminus = {x}^{3} \ln x - {x}^{3} / 2 + C x$