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

The solution for y^{''}-6y^'+25y=e^t is y=e^{3t}(c_{1}cos(4t)+c_{2}sin(4t))+1/20 e^t

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

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

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{1}{20} e^{t}

Solution steps

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

Solve linear ODE:

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{1}{20} e^{t}

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{1}{20} e^{t}

\int x^{3} arctan (x) d x

\frac{d}{d θ} (θ sin (θ))

x^{2} + 3 x - 2

\frac{d ^{3}}{d x ^{3}} (arctan (x))

- 0.1 x^{2} - x + 40

What is the solution for y^{''}-6y^'+25y=e^t ?
The solution for y^{''}-6y^'+25y=e^t is y=e^{3t}(c_{1}cos(4t)+c_{2}sin(4t))+1/20 e^t