Differential Equations/differntial equation question
(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...