What is the solution for 3x^{''}+25x^'+48x=5cos(4t) ?

The solution for 3x^{''}+25x^'+48x=5cos(4t) is x=c_{1}e^{-3t}+c_{2}e^{-(16t)/3}+1/20 sin(4t)

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3x^{''}+25x^'+48x=5cos(4t)

3 x^{^{''}} + 25 x^{^{'}} + 48 x = 5 cos (4 t)

x = c_{1} e^{- 3 t} + c_{2} e^{- \frac{16 t}{3}} + \frac{1}{20} sin (4 t)

Solution steps

3 x^{''} (t) + 25 x^{'} (t) + 48 x = 5 cos (4 t)

Solve linear ODE:

x = c_{1} e^{- 3 t} + c_{2} e^{- \frac{16 t}{3}} + \frac{1}{20} sin (4 t)

x = c_{1} e^{- 3 t} + c_{2} e^{- \frac{16 t}{3}} + \frac{1}{20} sin (4 t)

What is the solution for 3x^{''}+25x^'+48x=5cos(4t) ?
The solution for 3x^{''}+25x^'+48x=5cos(4t) is x=c_{1}e^{-3t}+c_{2}e^{-(16t)/3}+1/20 sin(4t)