Linear diff eq - correctly done?

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Asphyxiated
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Homework Statement



Solve the linear differential equation:

[tex]xy'-2y=x^{2}[/tex]


Homework Equations



If you have a linear differential equation of the form:

[tex]y'+P(x)y=Q(x)[/tex]

then your integrating factor is:

[tex]I(x)=e^{\int P(x) dx}[/tex]

The Attempt at a Solution



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

[tex]y' - \frac {2y}{x} = x[/tex]

where

[tex]P(x)=-\frac{2}{x}[/tex]

thus:

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

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

[tex]x^{-2}y'-2x^{-3}y=\frac{1}{x}[/tex]

which is:

[tex]\frac {d}{dx}(x^{-2}y)=\frac{1}{x}[/tex]

if we the integrate both sides:

[tex]x^{-2}y=ln|x|+c[/tex]

therefore:

[tex]y= x^{2}ln|x|+cx^{2}[/tex]
 
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Looks good to me.
 
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.