# differential eqWatch

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#1
x^2 y'' - 2xy' + 2y = Ln(x), y(1)=y'(1)=1

i cant set it up. can someone get me started with this Q. not do it but jus show me the initial step
0
14 years ago
#2
substuting x=1 into the eq, then we find y''(1)=0
0
14 years ago
#3
Okay, it's been a while since I did these but here goes.

Use the substitution z=lnx

Then we get dz/dx = 1/x and d^2z/dx^2 = -1/x^2

Now by the chain rule we know that

dy/dx = dy/dz * dz/dx

And differentiating with respect to x gives us

d^2y/dx^2 = dz/dx * d^2z/dx^2 + (dz/dx)^2 * d^2y/dz^2

Once we substitute this all into the original equation we get

- d^2y/dz^2 - dy/dz + 2y = z

Which you should be able to solve easily.
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