請幫我解決工程數學問題! 謝謝?

1. y'+2y=4e^x

P(x)=2
I=e^∫P(x)dx=e^2x

(ye^(2x))'= 4e^(3x)
ye^(2x) = ∫4e^(3x) dx
ye^(2x) = (4/3)e^(3x) + C for C is a constant
y = (4/3)e^x + Ce^(-2x)
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2. y''+3y'+2y=2x²

y''+3y'+2y=0..............homo
Let y = e^(rx)
y'=re^(rx)
y''=(r^2)e^(rx)

(e^(rx))(r^2+3r+2)=0
(r+1)(r+2)=0
r=-1 or -2

y_c(x)=C1e^(-x) + C2e^(-2x) for C1 and C2 are constant..............homo sol



y''+3y'+2y=2x²..............particular
Let Y = Ax^2 + Bx + C
Y' = 2Ax + B
Y'' = 2A

input Y, Y', and Y'' into y''+3y'+2y=2x² we get
2A + 3(2Ax+B) + 2(Ax^2 + Bx + C) = 2x^2
解放程式
得到
A=1/2
B=-3/2
C=7/4

因此得到
Y = Ax^2 + Bx + C
Y = (1/2)x^2 + (-3/2)x + 7/4..............particular sol


最後
把homo sol加上particular sol得到
y(x) = y_c(x) + Y
y(x) = C1e^(-x) + C2e^(-2x) + (1/2)x^2 + (-3/2)x + 7/4

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3. y''+2y'-3y=e^x

y''+2y'-3y=0..............homo
Let y = e^(rx)
y'=re^(rx)
y''=(r^2)e^(rx)

(e^(rx))(r^2+2r-3)=0
(r^2+2r-3)=0
(r+3)(r-1)=0
r=-3 or 1

y_c(x)=C1e^(-3x) + C2e^(x) for C1 and C2 are constant..............homo sol



y''+2y'-3y=e^x ..............particular
Let Y = Axe^x
Y' = Ae^x + Axe^x
Y'' = 2Ae^x + Axe^x

input Y, Y', and Y'' into y''+2y'-3y=e^x we get
(2Ae^x + Axe^x) + 2(Ae^x + Axe^x) - 3(Axe^x) = e^x
4Ae^x = e^x
解放程式
得到
A=1/4

因此得到
Y = Axe^x
Y = (1/4)xe^x..............particular sol


最後
把homo sol加上particular sol得到
y(x) = y_c(x) + Y
y(x) = C1e^(-3x) + C2e^(x) + (1/4)xe^x

謝謝大大