What is the solution for y^{''}-2y^'=7cos(3t) ?

The solution for y^{''}-2y^'=7cos(3t) is y=c_{1}e^{2t}+c_{2}-14/39 sin(3t)-7/13 cos(3t)

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y^{''}-2y^'=7cos(3t)

y^{^{''}} - 2 y^{^{'}} = 7 cos (3 t)

y = c_{1} e^{2 t} + c_{2} - \frac{14}{39} sin (3 t) - \frac{7}{13} cos (3 t)

Solution steps

y^{''} (t) - 2 y^{'} (t) = 7 cos (3 t)

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} - \frac{14}{39} sin (3 t) - \frac{7}{13} cos (3 t)

y = c_{1} e^{2 t} + c_{2} - \frac{14}{39} sin (3 t) - \frac{7}{13} cos (3 t)

y^{^{''}} - 16 y^{^{'}} + 64 y = 504 sin (x) - 128 cos (x)

y^{^{''''}} - 3 y^{^{'''}} + 2 y^{^{''}} = e^{3 x}

y^{^{''}} + y^{^{'}} = - cos (x)

y^{^{''}} + 4 y = x sin (x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 e^{- x}

What is the solution for y^{''}-2y^'=7cos(3t) ?
The solution for y^{''}-2y^'=7cos(3t) is y=c_{1}e^{2t}+c_{2}-14/39 sin(3t)-7/13 cos(3t)