What is the solution for 3x^{''}+48x=5cos(4t) ?

The solution for 3x^{''}+48x=5cos(4t) is x=c_{1}cos(4t)+c_{2}sin(4t)+(5tsin(4t))/(24)

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3x^{''}+48x=5cos(4t)

3 x^{^{''}} + 48 x = 5 cos (4 t)

x = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{5 t sin ( 4 t )}{24}

Solution steps

3 x^{''} (t) + 48 x = 5 cos (4 t)

Solve linear ODE:

x = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{5 t sin ( 4 t )}{24}

x = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{5 t sin ( 4 t )}{24}

8 y^{^{''}} + 6 y^{^{'}} - y = 11

y^{^{''}} - 6 y^{^{'}} = 37 sin (x)

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 e^{- 2 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + y = 20 sin (x)

y^{^{''}} + 4 y = cos (3 x)

What is the solution for 3x^{''}+48x=5cos(4t) ?
The solution for 3x^{''}+48x=5cos(4t) is x=c_{1}cos(4t)+c_{2}sin(4t)+(5tsin(4t))/(24)