What is the solution for y^{''}+2y^'=2x+5-e-2x ?

The solution for y^{''}+2y^'=2x+5-e-2x is y=c_{1}+c_{2}e^{-2x}+((-e+5)x)/2

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y^{''}+2y^'=2x+5-e-2x

y^{^{''}} + 2 y^{^{'}} = 2 x + 5 - e - 2 x

y = c_{1} + c_{2} e^{- 2 x} + \frac{( - e + 5 ) x}{2}

Solution steps

y^{''} (x) + 2 y^{'} (x) = 2 x + 5 - e - 2 x

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{( - e + 5 ) x}{2}

y = c_{1} + c_{2} e^{- 2 x} + \frac{( - e + 5 ) x}{2}

y^{^{''}} - 2 y^{^{'}} + y = x^{3} + 8 x

y^{^{''}} - 9 y = (x^{2} - 8) sin (3 x)

y^{^{''}} - 3 y^{^{'}} = sin (4 t)

y^{^{''}} + 4 y = 2 x, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + a y^{^{'}} + b y = x, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+2y^'=2x+5-e-2x ?
The solution for y^{''}+2y^'=2x+5-e-2x is y=c_{1}+c_{2}e^{-2x}+((-e+5)x)/2