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(x^2)y''+xy'+y=x with y(1)=y'(1)=1. Then what is the value of y(e^π/2)

First solve the associated homogeneous problem: (x^2)y''+xy'+y=0, by letting y=x^n

Substituting that into the DE gives: (x^n)*(n^2+1)=0 ==> n=-i or +i.

Thus, y1=x^i or y2=x^(-i). Using Euler's formula e^(bi)=cos(b)+isin(b), we get

y1=cos(ln(x))+isin(ln(x)), and y2=cos(ln(x))-isin(ln(x)). Combining them as a

linear combination to obtain y(x), gives y(x)=k1*y1+k2*y2= (k1+k2)cos(ln(x))+(k1-k2)isin(ln(x))

Letting C1=k1+k2 and C2=(k1-k2)i, gives the homogeneous soluiton: y(x)=C1*cos(ln(x))+C2*sin(ln(x)).

Now for the particular solution to (x^2)y''+xy'+y=x

Solve by Variation of Parameters, to obtain y(x)=(1/2)x

Thus, the complete solution is y(x)=C1*cos(ln(x))+C2*sin(ln(x))+(1/2)x.

Now you can substitute the IC's in to C1=1/2 and C2=1/2....then let x=e^n/2...

OK, good?

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Differential Equations

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Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

Over 15 years teaching at the college level.**Organizations**

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B.S. in Mathematics from Rensselaer Polytechnic Institute

M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook