Answer to Question #148540 in Differential Equations for Nikhil

Question

Solve the differential equation
xdy/dx +y = y^2Inx

Answer 1

Given differential equation: $x\frac{dy}{dx}+y=y^2lnx$

Dividing the given equation by $x$ , we get

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

Observe that the above differential equation is Bernoulli's differential equation.

It can be reduced to Linear Differential equation dividing the equation by $y^2$ and then take the substitution $\frac{1}{y}=z\Rightarrow \frac{1}{y^2}\frac{dy}{dx}=-\frac{dz}{dx}$

Then the given differential equation becomes

$\frac{1}{y^2}\frac{dy}{dx}+\frac{1}{y}\frac{1}{x}=\frac{lnx}{x}$

$\Rightarrow -\frac{dz}{dx}+\frac{z}{x}=\frac{lnx}{x} \Rightarrow \frac{dz}{dx}-\frac{z}{x}=-\frac{lnx}{x}$

The above is a first order linear differential equation of the form

$\frac{dz}{dx}+p(x)z=g(x)$ , where $p(x)=-\frac{1}{x}$ and $g(x)=-\frac{lnx}{x}$

General solution to the differential equation is

$z$ (Integrating Factor)$=\int$ (Integrating Factor)($g(x)dx+C$ , where

Integrating Factor $=e^{\int p(x)dx}$

Using the above, we have

Integrating Factor = $=e^{\int (-\frac{1}{x})dx}=e^{-lnx}=e^{ln(\frac{1}{x})}=\frac{1}{x}$

General solution is

$z(\frac{1}{x})=\int (\frac{1}{x})(-\frac{lnx}{x})dx+C=-\int \frac{lnx}{x^2}dx+C$

Take the substitution $lnx=u\Rightarrow \frac{1}{x}dx=du$ and $\frac{1}{x}=e^{-u}$

Then, we get

$\frac{z}{x}=-\int (ue^{-u})du+C=-[-ue^{-u}-e^{-u}]+C$ (Using integration by parts)

That is we get

$\frac{z}{x}=[ue^{-u}+e^{-u}]+C=\frac{lnx}{x}+\frac{1}{x}+C$ $(u=lnx)$

That is we get

$\frac{1}{xy}=\frac{lnx}{x}+\frac{1}{x}+C(z=\frac{1}{y})$

Therefore, general solution to the given differential equation is

$ylnx+y+C=1$