What is the solution for y^{''}+4y^'+5y=3t+e^t ?

The solution for y^{''}+4y^'+5y=3t+e^t is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+(3t)/5-12/25+1/10 e^t

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y^{''}+4y^'+5y=3t+e^t

y^{^{''}} + 4 y^{^{'}} + 5 y = 3 t + e^{t}

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{3 t}{5} - \frac{12}{25} + \frac{1}{10} e^{t}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 3 t + e^{t}

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{3 t}{5} - \frac{12}{25} + \frac{1}{10} e^{t}

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{3 t}{5} - \frac{12}{25} + \frac{1}{10} e^{t}

y^{^{''}} - 25 y = 3 x^{2} - 2 + e^{x}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 4 y = 8 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} + 40 y = 25 sin (6 t)

x^{2} y^{^{''}} - 8 x^{2} y^{^{'}} + 16 x^{2} y = 3 e^{4 x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 e^{t}, y (0) = 2, y^{^{'}} (0) = 3

What is the solution for y^{''}+4y^'+5y=3t+e^t ?
The solution for y^{''}+4y^'+5y=3t+e^t is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+(3t)/5-12/25+1/10 e^t