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Name: Find the general solution of y″−2y′+2y=xexcos⁡x
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find the general solution of y″−2y′+2y=xexcos⁡x

Tyra

Answered question

2020-10-20

Find the general solution of

y^{″} - 2 y^{'} + 2 y = x e^{x} \cos x

Answer & Explanation

Aniqa O'Neill

Skilled2020-10-21Added 100 answers

y^{″} - 2 y^{'} + 2 y = x e^{x} \cos (x) .

Let

s = \sin (x), c = \cos (x)

, s'=c, c'=-s, then:

y^{″} - 2 y^{'} + 2 y = x e^{x} c

.
First solve the homogeneous equation:

y^{″} - 2 y^{'} + 2 y = 0 .

The characteristic equation is

r^{2} - 2 r + 2 = 0 = (r - 1 + i) (r - 1 - i) .

y = A e^{(1 + i) x} + B e^{(1 - i) x}

, which can be written:

y = e^{x} (A e^{i x} + B e^{- i x}),

where A and B are constants.
Use Euler’s expansion:

y = e^{x} (A \cos (x) + A i \sin (x) + B \cos (x) - B i \sin (x)),

y = e^{x} ((A + B) \cos (x) + i (A - B) \sin (x)) .

Let C=A+B and D=A-B, then:

y = e^{x} (C \cos (x) + D \sin (x)) .

Call this general solution

y_{1} = e^{x} (C \cos (x) + D \sin (x)) = e^{x} (C c + D s)

Now we have to deal with xeˣc, the particular solution, which will give us y₂, such that the entire solution is

y = y_{1} + y_{2}

.
We guess the solution, where P, Q, R and S are constants:
Let

u = (P x + Q) s + (R x + S) c

and apply the tentative solution

y_{2} = x e^{x} u

u^{'} = (P x + Q) c + P s - (R x + S) s + R c,

u^{″} = - (P x + Q) s + 2 P c - (R x + S) c - 2 R s,

y_{2}^{'} = x e^{x} u^{'} + u e^{x} (x + 1),

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