What is the solution for y^{''}-3y^'+2y=3cos(3t) ?

The solution for y^{''}-3y^'+2y=3cos(3t) is y=c_{1}e^{2t}+c_{2}e^t-27/130 sin(3t)-21/130 cos(3t)

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y^{''}-3y^'+2y=3cos(3t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 3 cos (3 t)

y = c_{1} e^{2 t} + c_{2} e^{t} - \frac{27}{130} sin (3 t) - \frac{21}{130} cos (3 t)

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = 3 cos (3 t)

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} - \frac{27}{130} sin (3 t) - \frac{21}{130} cos (3 t)

y = c_{1} e^{2 t} + c_{2} e^{t} - \frac{27}{130} sin (3 t) - \frac{21}{130} cos (3 t)

y^{^{''}} - 12 y^{^{'}} + 36 y = 72 x + 225 cos (3 x)

y^{^{''}} + a \cdot y^{^{'}} + b^{2} \cdot y = cos (b \cdot x)

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

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{- 3 t}

y^{^{''}} + 4 y^{^{'}} + 4 y = 16 t + 13

What is the solution for y^{''}-3y^'+2y=3cos(3t) ?
The solution for y^{''}-3y^'+2y=3cos(3t) is y=c_{1}e^{2t}+c_{2}e^t-27/130 sin(3t)-21/130 cos(3t)