Find the general solution of the equation?

Question

(1-x^2)(d^2y/dx^2) = x(a-y)
Thanks!!

Answer 1

The previous answerer's method is incorrect.

For example, consider y = x e^x; y' = e^x + xe^x and y" = 2e^x + xe^x = y(1 + 2/x).
If we write y"/y = 1 + 2/x and integrate both sides twice, we get
∫ ln y = ∫ (x + 2 ln x + c)
=> y (ln y - 1) = x^2 / 2 + 2x (ln x - 1) + cx + d
= (1/2) x^2 + 2 x ln x + cx + d (absorbing -2x into cx).

However, the LHS corresponds to xe^x (ln x + x - 1). No choice of c and d will make this work!

The separation technique is basically only valid for first order ODEs. To see this, suppose that F is an antiderivative of f. Then d/dx(F(y)) = (dF/dy) (dy/dx) by the chain rule = f(y) dy/dx. So if we have an ODE of the form f(y) dy/dx = g(x), and G is an antiderivative of G, we obtain F(y) = G(x) + c. So we can convert f(y) dy/dx = g(x) into ∫f(y) dy = ∫g(x) dx.

However, this doesn't work for second derivatives:
d2/dx^2 (F(y)) = d/dx ([dF/dy][dy/dx])
= ((d2F/dy^2) (dy/dx) (dy/dx) + (dF/dy)(d^2y/dx^2))
= f'(y) (dy/dx)^2 + f(y) (d2y/dx^2)
So we need the equation to be in this rather special form to do the equivalent trick.

Unfortunately, I haven't been able to find an analytic solution to your equation - though I am a bit rusty on these, so one may exist. Obviously y=a is a particular solution. 8-)

Answer 2

Separate variable

[1/(a-y)] d^2y = [x/(1-x^2)] dx^2

and integrate twice

Ana