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)

y^{^{''}} - 9 y = 24

\frac{d ^{2} y}{d t ^{2}} - 6 \frac{d y}{d t} + 9 y = t

y^{^{'''}} + 9 y^{^{'}} = 27 x^{2}

4 y^{^{''}} - y = e^{\frac{x}{2}}

y^{^{'''}} + 9 y^{^{''}} + 27 y^{^{'}} + 27 y = 5 e^{- 3 x}

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)