What is the solution for x^{''}+25x=32cos(3t) ?

The solution for x^{''}+25x=32cos(3t) is x=c_{1}cos(5t)+c_{2}sin(5t)+2cos(3t)

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x^{''}+25x=32cos(3t)

x^{^{''}} + 25 x = 32 cos (3 t)

x = c_{1} cos (5 t) + c_{2} sin (5 t) + 2 cos (3 t)

Solution steps

x^{''} (t) + 25 x = 32 cos (3 t)

Solve linear ODE:

x = c_{1} cos (5 t) + c_{2} sin (5 t) + 2 cos (3 t)

x = c_{1} cos (5 t) + c_{2} sin (5 t) + 2 cos (3 t)

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 6 y = 6

2 y^{^{''}} + y^{^{'}} + 3 y = 3 cos (3 t) + 2 sin (3 t)

y^{^{''}} - y = x^{2}, y (0) = - 3, y^{^{'}} (0) = 5

x^{^{''}} - 6 x^{^{'}} + 13 x = 2197 t^{2}

(d^{2} + 1) y = e^{x}

What is the solution for x^{''}+25x=32cos(3t) ?
The solution for x^{''}+25x=32cos(3t) is x=c_{1}cos(5t)+c_{2}sin(5t)+2cos(3t)