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

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

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y^{''}-4y^'+5y=5t

y^{^{''}} - 4 y^{^{'}} + 5 y = 5 t

y = e^{2 t} (c_{1} cos (t) + c_{2} sin (t)) + t + \frac{4}{5}

Solution steps

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

Solve linear ODE:

y = e^{2 t} (c_{1} cos (t) + c_{2} sin (t)) + t + \frac{4}{5}

y = e^{2 t} (c_{1} cos (t) + c_{2} sin (t)) + t + \frac{4}{5}

\frac{d ^{2} y}{d x ^{2}} + y = x

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

y^{^{''}} + 9 y = \frac{1}{( cos ( 3 x ) )}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 25 y = 2

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

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