What is the solution for y^{''}+2y^'=2x+3-e-2x ?

The solution for y^{''}+2y^'=2x+3-e-2x is y=c_{1}+c_{2}e^{-2x}+((-e+3)x)/2

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y^{''}+2y^'=2x+3-e-2x

y^{^{''}} + 2 y^{^{'}} = 2 x + 3 - e - 2 x

y = c_{1} + c_{2} e^{- 2 x} + \frac{( - e + 3 ) x}{2}

Solution steps

y^{''} (x) + 2 y^{'} (x) = 2 x + 3 - e - 2 x

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{( - e + 3 ) x}{2}

y = c_{1} + c_{2} e^{- 2 x} + \frac{( - e + 3 ) x}{2}

y^{^{''}} + 16 y = 7 sec (4 x)

y^{^{''}} + y^{^{'}} + 2 y = - 25 x e^{x}, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 13 = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{x} + cos (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e x - 12

What is the solution for y^{''}+2y^'=2x+3-e-2x ?
The solution for y^{''}+2y^'=2x+3-e-2x is y=c_{1}+c_{2}e^{-2x}+((-e+3)x)/2