What is the solution for y^{''}-6y^'+25y=64e^{-t} ?

The solution for y^{''}-6y^'+25y=64e^{-t} is y=e^{3t}(c_{1}cos(4t)+c_{2}sin(4t))+2e^{-t}

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y^{''}-6y^'+25y=64e^{-t}

y^{^{''}} - 6 y^{^{'}} + 25 y = 64 e^{- t}

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 2 e^{- t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 25 y = 64 e^{- t}

Solve linear ODE:

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 2 e^{- t}

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 2 e^{- t}

\int 2 (x + 2)^{3} d x

\frac{\partial}{\partial x} (x^{3} + 3 x y^{2} - 15 x - 12 y)

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} - 24 y = 0

x \to 0 - lim (x - 1)

\int_{0}^{1} 3 (x^{2} + 1) e^{- x} d x

What is the solution for y^{''}-6y^'+25y=64e^{-t} ?
The solution for y^{''}-6y^'+25y=64e^{-t} is y=e^{3t}(c_{1}cos(4t)+c_{2}sin(4t))+2e^{-t}