solve (x^2-1)y''-2xy'+2y=(x^2-1)^2
this does not correspond to any form how to solve it
That is a "linear differential equation with variable coefficients". It has x= 1 as a "regular singular value" and you need to use "Frobenius's method" to find a solution about x= 1 (if, for example, the problem includes initial values at x= 1). For solutions about x, not equal to 1, you use a regular power series solution.
Hi prasum !
If you have solved the homogeneous ODE : (x²-1)y''-2xy'+2y=0, you obtained :
y = c1*x+c2(x+1)²
Then, in order to solve the complete ODE : (x²-1)y''-2xy'+2y=(x²-1)² , you have to find a particular solution.
Consider one solution of the homogeneous ODE and replace its coefficient by an unknown function f(x). For example :
y = f(x)*x (which is simpler than the other possible option f(x)*(x+1)²)
Compute f'(x) and f''(x) and bring them back into (x²-1)y''-2xy'+2y=(x²-1)²
You will observe that the the second order ODE is reduced to a first order ODE with the unknown g(x)=f'(x).
This linear first order ODE is easy to solve, leading to g(x) then, after integration, to f(x) and finally to y(x)=x*f(x).
You may get an exact solution:
Let $\displaystyle y_p(x)$ a particular solution, $\displaystyle y_p(x) = x^2+1$. Because the differential equation is a linear one, we may use d'Alambert's reduction method: $\displaystyle y(x)=y_p(x)u(x)$. If we substitute it to the homogenous equation for $\displaystyle y(x)$, then we get:
$\displaystyle (x^4-1)u''+2x(x^2-3)u'=0$.
After substituting $\displaystyle u'(x)=v(x)$, we may conclude to a separable first order ODE. We then solve it for $\displaystyle v$, integrate it to get $\displaystyle u$ and finally $\displaystyle y(x)=(x^2+1)u(x)$ will be the solution.