Why Is My Solution Different from the Textbook's for This Linear ODE?

[verd]

Hey,

So I solved this one linear DE and the answer I got isn't the same as the one in the back of my textbook... And I'm not sure why. I thought I was doing this right. Could someone tell me what I'm doing wrong?

Here's what I'm given:

[tex]x^{2}y'+xy=1[/tex]
[tex]\frac{dy}{dx}+\frac{x}{x^{2}}y=\frac{1}{x^{2}}[/tex]
[tex]\frac{dy}{dx}+\frac{1}{x}y=\frac{1}{x^{2}}[/tex]
[tex]integrating-factor-p=\frac{1}{x}[/tex]
[tex]integrating-factor\rightarrow e^{\int p(x) dx}=e^{\int \frac{1}{x}dx}= e^{lnx}=x[/tex]
[tex](xy)'=\frac{1}{x^{2}}[/tex]
[tex]xy=\int x^{-2}dx=\frac{-1}{x}+C[/tex]
[tex]y=\frac{-1}{x^{2}}+\frac{C}{x}[/tex]

That's what I get.

The answer in the textbook is:
[tex]y=x^{-1}lnx+Cx^{-1}[/tex]

Which is similar but not the same... What'd I do wrong?

 

[EbolaPox]

When you get an integrating factor, multiply it through both sides to obtain

[tex] (xy)' = \frac{1}{x} [/tex]
Then, Integrate to obtain
[tex] \int d(xy) = \int \frac{1}{x} dx [/tex]
And this leads you to
[tex] y = \frac{ln(x)}{x} + Cx^{-1} [/tex]

The step where you made the error was after the integrating factor, you did not also apply it to the right side of the equation, the [tex] x^{-2} [/tex].