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

The solution for y^{''}+3y^'-4y=cos(3t) is y=c_{1}e^t+c_{2}e^{-4t}+9/250 sin(3t)-13/250 cos(3t)

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y^{''}+3y^'-4y=cos(3t)

y^{^{''}} + 3 y^{^{'}} - 4 y = cos (3 t)

y = c_{1} e^{t} + c_{2} e^{- 4 t} + \frac{9}{250} sin (3 t) - \frac{13}{250} cos (3 t)

Solution steps

y^{''} (t) + 3 y^{'} (t) - 4 y = cos (3 t)

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 4 t} + \frac{9}{250} sin (3 t) - \frac{13}{250} cos (3 t)

y = c_{1} e^{t} + c_{2} e^{- 4 t} + \frac{9}{250} sin (3 t) - \frac{13}{250} cos (3 t)

What is the solution for y^{''}+3y^'-4y=cos(3t) ?
The solution for y^{''}+3y^'-4y=cos(3t) is y=c_{1}e^t+c_{2}e^{-4t}+9/250 sin(3t)-13/250 cos(3t)