differential equation. Please help!!!?

2017-10-14 11:46 pm

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?

更新1:

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.

回答 (1)

Lopez

2017-10-15 3:48 pm

✔ 最佳答案

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

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