x^2 y'' + 4xy' - 10y = 3/x^5 , x > 0

aldo101

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May 20, 2007
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I came across this prob. And I can't solve it. If somebody can please help me. Thank you.

x^2 y'' + 4xy' - 10y = 3/x^5 , x > 0
 
You can use Variation of Parameters.

The auxiliary equation is \(\displaystyle \L\\m(m-1)+4m-10=(m-2)(m+5)=0\)

Thus, \(\displaystyle \L\\y_{c}=C_{1}x^{2}+C_{2}x^{-5}\)

Divide the equation by x^2 and get: \(\displaystyle \L\\y''+\frac{4}{x}y'-\frac{10}{x^{2}}y=\frac{3}{x^{7}}\)

Make the identification \(\displaystyle \L\\f(x)=\frac{3}{x^{7}}\).

Now with \(\displaystyle \L\\y_{1}=x^{2} \;\ and \;\ y_{2}=x^{-5}\), build the Wronskians:

\(\displaystyle \L\\W, \;\ W_{1}, \;\ W_{2}, \;\ u'_{1}, \;\ u'_{2}\)

Write back if you remain stuck.
 
I got it. I wasn't suer if variation was going to work. I see tht it does. Thank you.
 
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