Linear diff eq - correctly done?

[Asphyxiated]

Homework Statement

Solve the linear differential equation:

xy'-2y=x^{2}

Homework Equations

If you have a linear differential equation of the form:

y'+P(x)y=Q(x)

then your integrating factor is:

I(x)=e^{\int P(x) dx}

The Attempt at a Solution

If we divide both sides by x then the equation is in standard form:

y' - \frac {2y}{x} = x

where

P(x)=-\frac{2}{x}

thus:

I(x)=e^{\int -\frac{2}{x} dx} = e^{-2\int\frac{1}{x}dx}= e^{-2ln|x|}=x^{-2}

so we then multiple both sides of the diff eq by I(x):

x^{-2}y'-2x^{-3}y=\frac{1}{x}

which is:

\frac {d}{dx}(x^{-2}y)=\frac{1}{x}

if we the integrate both sides:

x^{-2}y=ln|x|+c

therefore:

y= x^{2}ln|x|+cx^{2}

 

[Char. Limit]

Looks good to me.

 

[Mark44]

Your solution is easy enough to check. Substitute your solution for y and take its derivative. When you evaluate xy' - 2y, you should get x².