- #1
Lancelot59
- 646
- 1
I've picked up a bit more since my last problem. I need to solve the following DE:
[tex]x^{2}\frac{dy}{dx}+x(x+2)y=e^{x}[/tex]
I decided to use variation of parameters, so I re-arranged it like so:
[tex]\frac{dy}{dx}=\frac{e^{x}}{x^{2}}-(1+\frac{2}{x})y[/tex]
Then solved the homogenous DE:
[tex]\frac{dy}{dx}=-(1+\frac{2}{x})y[/tex]
[tex]y=e^{-x}x^{-2}c[/tex]
Now for the particular solution:
[tex]y_{p}=u(x)e^{-x}x^{-2}c
[tex]\frac{dy}{dx}=u'(x)e^{-x}x^{-2}-u(x)e^{-x}x^{-2}-2u'(x)e^{-x}x^{-2}[/tex]
When I shoved this back in I wound up with this for u'(t):
[tex]u'(x)=e^{x}x^{-2}[/tex]
It seems...a bit strange. Did I mess up somewhere? It's a bit hard to integrate. I've gone over this several times already.
[tex]x^{2}\frac{dy}{dx}+x(x+2)y=e^{x}[/tex]
I decided to use variation of parameters, so I re-arranged it like so:
[tex]\frac{dy}{dx}=\frac{e^{x}}{x^{2}}-(1+\frac{2}{x})y[/tex]
Then solved the homogenous DE:
[tex]\frac{dy}{dx}=-(1+\frac{2}{x})y[/tex]
[tex]y=e^{-x}x^{-2}c[/tex]
Now for the particular solution:
[tex]y_{p}=u(x)e^{-x}x^{-2}c
[tex]\frac{dy}{dx}=u'(x)e^{-x}x^{-2}-u(x)e^{-x}x^{-2}-2u'(x)e^{-x}x^{-2}[/tex]
When I shoved this back in I wound up with this for u'(t):
[tex]u'(x)=e^{x}x^{-2}[/tex]
It seems...a bit strange. Did I mess up somewhere? It's a bit hard to integrate. I've gone over this several times already.
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