匿名使用者
匿名使用者 發問時間: 科學數學 · 3 年前

differential equation. Please help!!!?

Show by differentiation and substitution together with the Leibnitz theorem that the differential equation:

(4x^2)(d^2y/dx^2) - 4x(dy/dx) + (4x^2+3)y = 0

has a solution of the form y(x) = (x^n)(sin x) , and find the value of n.

I can find that

(4x^2)(d^2y/dx^2) - 4x(dy/dx) + (4x^2+3)y = (2n-1)((x^(n)sinx)(2n-3) + 4x^(n+1)cosx)

but I don't know how to show (x^n)(sin x) is a solution and find n.

Can anyone help?

已更新項目:

Why ( 4n² - 8n + 3 )x^n sin x + ( 8n - 4 )x^(n+1)cos x can be directly equal to 0 without showing x^n sin x is a solution.

1 個解答

評分
  • Lopez
    Lv 5
    3 年前
    最佳解答

    Suppose y = x^n sin x

    y' = nx^(n-1)sin x + x^n cos x

    y''

    = n(n-1)x^(n-2)sin x + nx^(n-1)cos x + nx^(n-1) cos x - x^n sin x

    = n(n-1)x^(n-2)sin x + 2nx^(n-1) cos x - x^n sin x

    4x²y'' - 4xy' + ( 4x² + 3 )y

    = 4n(n-1)x^n sin x + 8nx^(n+1)cos x - 4x^(n+2)sin x - 4nx^n sin x - 4x^(n+1)cos x + 4x^(n+2)sin x + 3x^n sin x

    = ( 4n² - 8n + 3 )x^n sin x + ( 8n - 4 )x^(n+1)cos x

    = 0

    ( 4n² - 8n + 3 )x^n sin x = - ( 8n - 4 )x^(n+1)cos x

    ( 2n - 1 )( 2n - 3 )x^n sin x = - 4( 2n - 1 )x^(n+1) cos x

    For all x, the eq. holds when n = 1/2

    That is, the ODE has a solution of the following form

    y = x^(1/2) sin x = ( √x )sin x

    Q.E.D.

    Ans: n = 1/2

還有問題?馬上發問,尋求解答。