What is the solution for x^{''}+36x=100cos(4t) ?

The solution for x^{''}+36x=100cos(4t) is x=c_{1}cos(6t)+c_{2}sin(6t)+5cos(4t)

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x^{''}+36x=100cos(4t)

x^{^{''}} + 36 x = 100 cos (4 t)

x = c_{1} cos (6 t) + c_{2} sin (6 t) + 5 cos (4 t)

Solution steps

x^{''} (t) + 36 x = 100 cos (4 t)

Solve linear ODE:

x = c_{1} cos (6 t) + c_{2} sin (6 t) + 5 cos (4 t)

x = c_{1} cos (6 t) + c_{2} sin (6 t) + 5 cos (4 t)

y^{^{''}} - 5 y^{^{'}} = 8 e^{4 t} - 4 e^{- t}

4 y^{^{''}} + 4 y^{^{'}} + y = e^{- x} + sin (x)

y^{^{''}} + 3 y^{^{'}} = 18

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 7

\frac{d ^{2} y}{d x ^{2}} + y = 4 x

What is the solution for x^{''}+36x=100cos(4t) ?
The solution for x^{''}+36x=100cos(4t) is x=c_{1}cos(6t)+c_{2}sin(6t)+5cos(4t)