series solution of xy''-y'=0,y(0)=0,y'(0)=1

Question

this the question about series

Answer 1

Let z=y' then z'=y". Substituting for y' and y":

xz'-z=0, xz'=z,

xdz/dx=z,

dz/z=dx/x,

ln(z)=ln(x)+C which can be written ln(z)=ln(Ax) where A=ln(C).

z=Ax=dy/dx,

dy=Axdx, y=Ax²/2+B.

y'=Ax, y"=A.

xy"-y'=Ax-Ax=0✔️ Therefore y=Ax²/2+B. But A/2 is still a constant which we could call 'a'.

To find A (or 'a') and B let's use the given equations y(0)=0 so B=0 and y=ax²; y'(0)=1 but y'=2ax and y'(0)=0 no matter what a is.

The given initial conditions don't appear to work in this case.

Another way to solve this is to rewrite the equation: y"-y'/x=0 (dividing through by x).

Multiply through by 1/x: y"/x-y'/x²=0=d(y'/x)/dx=d(x^-1y')/dx. (The derivative of x^-1y'=x^-1y"-y'x^-2=y"/x-y'/x².)

Hence y'/x=A after integration, where A is an arbitrary constant, and y'=Ax (as before), dy/dx=Ax, so y=Ax²/2+B or y=ax²+B. When x=0, y=0 so B=0. And y'(0)=0 makes A and a indeterminate.