# Solve dy/dx+y/x=y² by bernollies equation ?

Jul 1, 2018

The general solution is $y = \frac{1}{- x \ln x + C x}$

#### Explanation:

The Bernouilli ODE is of the form

$y ' + p \left(x\right) y = q \left(x\right) {y}^{n}$

The general solution is obtained by substituting

$v = {y}^{1 - n}$

and solving

$\frac{1}{1 - n} v ' + p \left(x\right) v = q \left(x\right)$

Here,

The equation is

$\frac{\mathrm{dy}}{\mathrm{dx}} + \frac{y}{x} = {y}^{2}$

$p \left(x\right) = \frac{1}{x}$

$q \left(x\right) = 1$

$n = 2$

Divide both sides by ${y}^{2}$

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

Let $v = \frac{1}{y}$, $\implies$, $y v = 1$

Differentiating both side wrt $x$

$v \frac{\mathrm{dy}}{\mathrm{dx}} + y \frac{\mathrm{dv}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{v} \frac{\mathrm{dv}}{\mathrm{dx}}$

Substituting in $\left(1\right)$

$- {v}^{2} \frac{y}{v} \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{v}{x} = 1$

$\frac{\mathrm{dv}}{\mathrm{dx}} - \frac{v}{x} = - 1$...................$\left(2\right)$

The integrating factor is

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

Multiply $\left(2\right)$ by $I F$

$\frac{1}{x} \frac{\mathrm{dv}}{\mathrm{dx}} - \frac{v}{x} ^ 2 = - \frac{1}{x}$

$\frac{d}{\mathrm{dx}} \left(\frac{v}{x}\right) = - \frac{1}{x}$

Integrating both sides

$\frac{v}{x} = - \ln x + C$

$v = - x \ln x + C x$

Substituting back $v = \frac{1}{y}$

$\implies$, $\frac{1}{y} = - x \ln x + C x$

$y = \frac{1}{- x \ln x + C x}$

Jul 1, 2018

$y = \frac{1}{x \left(C - \ln \left\mid x \right\mid\right)}$

#### Explanation:

Bernoulli General Form:

• $y ' + P \left(x\right) y = Q \left(x\right) {y}^{n} q \quad \text{with } \left\{\begin{matrix}P = \frac{1}{x} \\ Q = 1 \\ n = 2\end{matrix}\right.$

Standard substitution:

• $z \left(x\right) = {y}^{1 - n} = \frac{1}{y} q \quad \therefore z ' = - \frac{1}{y} ^ 2 y '$

So the DE $y ' + \frac{y}{x} = {y}^{2}$ becomes:

$- {y}^{2} \setminus z ' + \frac{y}{x} = {y}^{2}$

$z ' - \frac{1}{x y} = - 1 \implies z ' - \frac{z}{x} = - 1$

This is now linear, and can be solved by Integration factor:

• $\exp \left(\int - \frac{1}{x} \mathrm{dx}\right) = \frac{1}{x}$

$\frac{1}{x} \left(z ' - \frac{z}{x}\right) = - \frac{1}{x}$

${\left(\frac{1}{x} z\right)}^{'} = - \frac{1}{x}$

$\frac{1}{x} z = - \ln \left\mid x \right\mid + C$

$z = x \left(C - \ln \left\mid x \right\mid\right) = \frac{1}{y}$

$y = \frac{1}{x \left(C - \ln \left\mid x \right\mid\right)}$