Hey guys. I'm new here. I've been trying to figure out how to solve this problem, and I'm still confused.(adsbygoogle = window.adsbygoogle || []).push({});

(-x^2 + 4x -3)* d^{2}y/dx^{2}- 2(x-2) * dy/dx + 6y = 0

y(-2) = 1

dy/dx(-2) = 0

I set y = [tex]\sum[/tex]a_{n}(x+2)^{n}(start at n=0, n goes to infinity)

dy/dx = [tex]\sum[/tex]a_{n}n(x+2)^{n-1}(start at n=0, n goes to infinity)

d^{2}y/dx^{2}= [tex]\sum[/tex]a_{n}n(n-1)(x+2)^{n-2}(start at n=0, n goes to infinity

So:

0 = 6[tex]\sum[/tex]a_{n}(x+2)^{n}- 2(x+2)[tex]\sum[/tex]a_{n}n(x+2)^{n-1}+ (-x^2 + 4x -3)[tex]\sum[/tex]a_{n}n(n-1)(x+2)^{n-2}

I can simplify this to:

0 = [tex]\sum[/tex]6a_{n}(x+2)^{n}- [tex]\sum[/tex]2a_{n}n(x+2)^{n}+ (-x^2 + 4x -3)[tex]\sum[/tex]a_{n}n(n-1)*(x+2)^{n-2}

This further simplifies to:

0 = [tex]\sum[/tex](6a_{n}-2a_{n}n)(x+2)^{n}+ (-x^2 + 4x -3)[tex]\sum[/tex]a_{n}n(n-1)*(x+2)^{n-2}

However I don't know how to further simplify it, and where to go from here. Am I on the right track? I believe the answer will look like this:

y = a_{1}+ a_{2}x + a_{3}x^{2}+ a_{4}x^{3}.............

If somebody could please help me out, I would very much appreciate it. Thank You

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# Solving differential equation (involves power series)

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