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)

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

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} - 8 y^{^{'}} + 20 y = 100 x^{2} - 39 x e^{x}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x} csc (x)

y^{^{''}} + y = sin (3 x)

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)