If y c1 c2 x2 is the general solution of xyy 0 find the v

If y = c_1 + c_2 x^2 is the general solution of xy\"-y\' = 0 find the values of c that satisfy the boundary conditions: y(0)=1 and y\'(1)=1

Solution

if y=c₁+c₂x²  is the general solution of xy^\'\'-y^\'=0

If y = y(x) is a solution of (H) and if C is any real number, then u(x) = Cy(x) is also a solution of (H). Proof Let y = y(x) be a solution of (H). Then y00(x) + p(x)y0 (x) + q(x)y(x)=0. Let C be any real number, and set u(x) = Cy(x). Then u(x) = Cy(x) u0 (x) = Cy0 (x) u00(x) = Cy00(x) Substituting u into (H), we get u00(x) + p(x) u0 (x) + q(x) u(x) = Cy00(x) + p(x)[Cy0 (x)] + q(x)[Cy(x)] = C[y00(x) + p(x) y0 (x) + q(x) y(x)] = C[0] = 0. Alternate Proof Consider the linear differential operator L[y] = y00 + p(x)y0 + q(x)y. Since y = y(x) is a solution of (H), L[y(x)] = 0. Since L is a linear operator, L[Cy(x)] = C L[y(x)] = C(0) = 0. Thus, u(x) = Cy(x) is a solution of (H)