For each of the following differential equations, determine the general or particular solution: 2xy′+y=y2log⁡x

Step 1
Given:
Differential equation
${\displaystyle x{y}^{\prime }+y=y^2\log x}$
To find :
General solution of the given Differential equation?
Step 2
Solution :
We have given
${\displaystyle x{y}^{\prime }+y=y^2\log x}$
The given DE equation is a Bernoulli equation of the form
${\displaystyle \frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)y^n}$
And we can solve this kind of equation by substituting
${\displaystyle v=y^{1-n}}$
So we have the equation ${\displaystyle x{y}^{\prime }+y=y^2\log x}$
and here the value of n is 22
So we will substitute
${\displaystyle v=y^{1-2}}$
${\displaystyle v=\frac{1}{y}}$
Now differentiate with respect to x
${\displaystyle \frac{dv}{dx}=\frac{-1}{y^2}\frac{dy}{dx}}$
${\displaystyle \frac{dy}{dx}=-y^2\frac{dv}{dx}}$
Now we can write the given equation as
${\displaystyle x\frac{dy}{dx}+y=y^2\log x}$
${\displaystyle \frac{dy}{dx}+\frac{y}{x}=\frac{y^2\log x}{x}}$
Now put the value of ${\displaystyle \frac{dy}{dx}=-y^2\frac{dv}{dx}}$
${\displaystyle -y^2\frac{dv}{dx}+\frac{y}{x}=\frac{y^2\log x}{x}}$
divide both the sides by ${\displaystyle -y^2}$
${\displaystyle \frac{dv}{dx}-\frac{1}{xy}=\frac{-\log x}{x}}$
Now put ${\displaystyle v=\frac{1}{y}}$
${\displaystyle \frac{dv}{dx}-\frac{v}{x}=\frac{-\log x}{x}}$
Step 3
We got the equation
${\displaystyle \frac{dv}{dx}-\frac{v}{x}=\frac{-\log x}{x}}$
Now we will find out the integrating factor for the above equation

This is helpful

0

Flag