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

The solution for 3x^{''}+3x=5cos(4t) is x=c_{1}cos(t)+c_{2}sin(t)-1/9 cos(4t)

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

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

x = c_{1} cos (t) + c_{2} sin (t) - \frac{1}{9} cos (4 t)

Solution steps

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

Solve linear ODE:

x = c_{1} cos (t) + c_{2} sin (t) - \frac{1}{9} cos (4 t)

x = c_{1} cos (t) + c_{2} sin (t) - \frac{1}{9} cos (4 t)

y^{^{''}} + \frac{1}{8} y^{^{'}} + y = 3 \cdot cos (x)

y^{^{''}} + 5 y^{^{'}} - 6 y = 21 e^{t}

y^{^{''''}} + 2 y^{^{''}} + y = cos (2 x) + 2

y^{^{''}} - 2 y^{^{'}} + 5 y = 10 sin (x)

y^{^{''}} + \frac{1}{8} y^{^{'}} + y = 3 cos (x)

What is the solution for 3x^{''}+3x=5cos(4t) ?
The solution for 3x^{''}+3x=5cos(4t) is x=c_{1}cos(t)+c_{2}sin(t)-1/9 cos(4t)