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Find the general solution of the differential equation y '' - 2Y + 2Y = e ^ xsinx,
R ^ 2-2r + 2 = 0, R1 = 1 + I, R2 = 1-i
So the second equation is y = e (c1cosx + c2sinx)
Next, find a special solution
Let Y1 = Xe ^ x (b1cosx + b2sinx) be substituted into the original equation
Get B1, B2
y=Y+y1
Finding the general solution of the second order linear homogeneous differential equation y "+ 2Y = 0
The characteristic equation is R ^ 2 + 2 = 0
How to calculate the characteristic root is r = + / - root 2I? Ask a friend who understands to help. It's better to have the operation process,
It should be
The characteristic equation of the differential equation y "+ 2Y = 0 is R & sup2; + 2 = 0
∴r=±√2i
So the general solution of the differential equation y "+ 2Y = 0 is:
Y = c1cos (√ 2x) + c2sin (√ 2x), (C1, C2 are integral constants)
General solution of homogeneous linear ordinary differential equation y '' + 2Y '' + y = 0
The characteristic equation is: λ ^ 4 + 2, λ ^ 2 + 1 = 0
The solution is: λ = ± I (double root)
The special solutions are: cosx, SiNx, xcosx, xsinx
So the general solution is: y = C1 * cosx + c2sinx + C3 * xcosx + C4 * xsinx
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