What is the solution for 3x^{''}+53x=5cos(4t) ?

The solution for 3x^{''}+53x=5cos(4t) is x=c_{1}cos(sqrt(53/3)t)+c_{2}sin(sqrt(53/3)t)+cos(4t)

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3x^{''}+53x=5cos(4t)

3 x^{^{''}} + 53 x = 5 cos (4 t)

x = c_{1} cos (\frac{53}{3} t) + c_{2} sin (\frac{53}{3} t) + cos (4 t)

Solution steps

3 x^{''} (t) + 53 x = 5 cos (4 t)

Solve linear ODE:

x = c_{1} cos (\frac{53}{3} t) + c_{2} sin (\frac{53}{3} t) + cos (4 t)

x = c_{1} cos (\frac{53}{3} t) + c_{2} sin (\frac{53}{3} t) + cos (4 t)

y^{^{''}} - y^{^{'}} - 2 y = 3 cos (3 x)

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{a x}

25 y^{^{''}} - 10 y^{^{'}} - 2 y = 4 e^{t}

y^{^{''}} + 0.5 y^{^{'}} + y = cos (t)

y^{^{''}} + y = 2 cos (x), y (0) = 1, y^{^{'}} (0) = 0

What is the solution for 3x^{''}+53x=5cos(4t) ?
The solution for 3x^{''}+53x=5cos(4t) is x=c_{1}cos(sqrt(53/3)t)+c_{2}sin(sqrt(53/3)t)+cos(4t)