What is the solution for y^{''}+4y^'=10t+1 ?

The solution for y^{''}+4y^'=10t+1 is y=c_{1}+c_{2}e^{-4t}+(5t^2)/4-(3t)/8

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y^{''}+4y^'=10t+1

y^{^{''}} + 4 y^{^{'}} = 10 t + 1

y = c_{1} + c_{2} e^{- 4 t} + \frac{5 t ^{2}}{4} - \frac{3 t}{8}

Solution steps

y^{''} (t) + 4 y^{'} (t) = 10 t + 1

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} + \frac{5 t ^{2}}{4} - \frac{3 t}{8}

y = c_{1} + c_{2} e^{- 4 t} + \frac{5 t ^{2}}{4} - \frac{3 t}{8}

y^{^{''}} + y = cos (x) sec (x)

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

y^{^{''}} - 5 y^{^{'}} + 6 y = 5 x - 5

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 t e^{2 t}, y (0) = 1, y^{^{'}} (0) = 0

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

What is the solution for y^{''}+4y^'=10t+1 ?
The solution for y^{''}+4y^'=10t+1 is y=c_{1}+c_{2}e^{-4t}+(5t^2)/4-(3t)/8