Question

# Solve the differential equationx dy/dx-y= x^2lnx

First order differential equations

Solve the differential equation $$x dy/dx-y= x^2\ln x$$

2020-11-17

Divide both sides by x and further simplify it
$$x/x dy/dx-y/x= (x^2 \ln x)/x$$
$$dy/dx-y/x= x\ln x$$
It is of the form $$dy/dx+P(x)y= Q(x)$$
Now find the integrating factor by using $$(I.F.)= e^{\int P(x)dx}$$
Hence, $$I.F.= e^{-\int 1/x dx}$$
$$= e^{-\ln x}$$
$$= e^{\ln 1/x}$$
$$= 1/x$$
Hence, $$I.F.= 1/x$$
$$y*(I.F.)= \int (I.F.)*Q(x)dx+c$$
Hence, $$y*1/x= \int 1/x * x\ln x dx+c$$
$$y*1/x= \int \ln x dx+c$$
$$y*1/x= \int \ln x*1dx+c$$
$$y*1/x= \ln x+x+c$$
Simplify
$$y*1/x=x(\ln x+1)+c$$
$$y=x^2(\ln x+1)+cx$$