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Find the general solution of

$y'' + \dfrac{7}{x} y' + \dfrac{8}{x^2} y = 1, x > 0$

I don't even know how to solve the homogeneous version because it involves variables...

Does anyone know how to solve it?

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Yes, I missed that, edited –  paul marcus Feb 9 '13 at 3:53
    
Try $y=Ax^2$ to get a solution. –  Maesumi Feb 9 '13 at 3:54
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Then try $y=Cx^n$ for a homogeneous solution, here you find $n$. –  Maesumi Feb 9 '13 at 3:55
    
how can I find n? –  paul marcus Feb 9 '13 at 3:59
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The ode is known as "Euler Differential Equation". A related problem. –  Mhenni Benghorbal Feb 9 '13 at 5:43

2 Answers 2

$y=x^n$, $y'=nx^{n-1}$, $y''=n(n-1)x^{n-2}$. So $x^2y''+7xy'+8y=x^n[n(n-1)+7n+8]=0$ solve this equation for $n$.

So that gives $n^2+6n+8=0$ or $n=-4,-2$. The solution of homogeneous eq is $y_h=c_1x^{-4}+c_2x^{-2}$.

The particular solution here can be obtained by undetermined coefficient and guessing form of solution $y_p=Ax^2$. use that to solve $x^2y''+7xy'+8y=x^2$ here we get $A(2(2-1)+7*2+8)=1$ or $A=1/24$.

Now general solution is $y_g=y_h+y_p=c_1x^{-4}+c_2x^{-2}+x^2/24$.

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well, then n = -2 or n = -4, why is n = 2 also a solution? –  paul marcus Feb 9 '13 at 4:20
    
also, if I find two special solutions to the homogeneous solution, how can I get general solution to homogeneous one? Can I simply add c1, c2? –  paul marcus Feb 9 '13 at 4:28

It is of Euler differential equation type. Here is a related problem. You should have the following solution

$$ y(x) ={\frac {{\it c_2}}{{x}^{4}}}+{\frac {{\it c_1}}{{x}^{2}}}+\frac{{x}^{2}}{24}.$$

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