What is the solution for y^{''}+2y^'=2x+4 ?

The solution for y^{''}+2y^'=2x+4 is y=c_{1}+c_{2}e^{-2x}+(x^2)/2+(3x)/2

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y^{''}+2y^'=2x+4

y^{^{''}} + 2 y^{^{'}} = 2 x + 4

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e^{3 t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''''}} - y = 5 t + cos (t)

y^{^{''}} + y = t, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 6 y^{^{'}} + 34 = 0

What is the solution for y^{''}+2y^'=2x+4 ?
The solution for y^{''}+2y^'=2x+4 is y=c_{1}+c_{2}e^{-2x}+(x^2)/2+(3x)/2