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Math Help - solution of differential eqn

  1. #1
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    solution of differential eqn

    solve (x^2-1)y''-2xy'+2y=(x^2-1)^2

    this does not correspond to any form how to solve it
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  2. #2
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    Re: solution of differential eqn

    As usual, the first step consists in solving the related homogeneous ODE : (x^2-1)y''-2xy'+2y=0. What did you obtain ?
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  3. #3
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    Re: solution of differential eqn

    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.
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  4. #4
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    Re: solution of differential eqn

    Hi prasum !
    If you have solved the homogeneous ODE : (x-1)y''-2xy'+2y=0, you obtained :
    y = c1*x+c2(x+1)
    Then, in order to solve the complete ODE : (x-1)y''-2xy'+2y=(x-1) , 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))
    Compute f'(x) and f''(x) and bring them back into (x-1)y''-2xy'+2y=(x-1)
    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).
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  5. #5
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    Re: solution of differential eqn

    Didnt uNderstand
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  6. #6
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    Re: solution of differential eqn

    Did it and you will understand it.
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  7. #7
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    Re: solution of differential eqn

    You may get an exact solution:
    Let y_p(x) a particular solution, y_p(x) = x^2+1. Because the differential equation is a linear one, we may use d'Alambert's reduction method: y(x)=y_p(x)u(x). If we substitute it to the homogenous equation for y(x), then we get:
    (x^4-1)u''+2x(x^2-3)u'=0.
    After substituting u'(x)=v(x), we may conclude to a separable first order ODE. We then solve it for v, integrate it to get u and finally y(x)=(x^2+1)u(x) will be the solution.
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