What is the solution for y^{''}+6y^'+5y=10cos(5t) ?

The solution for y^{''}+6y^'+5y=10cos(5t) is y=c_{1}e^{-t}+c_{2}e^{-5t}+3/13 sin(5t)-2/13 cos(5t)

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y^{''}+6y^'+5y=10cos(5t)

y^{^{''}} + 6 y^{^{'}} + 5 y = 10 cos (5 t)

y = c_{1} e^{- t} + c_{2} e^{- 5 t} + \frac{3}{13} sin (5 t) - \frac{2}{13} cos (5 t)

Solution steps

y^{''} (t) + 6 y^{'} (t) + 5 y = 10 cos (5 t)

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 5 t} + \frac{3}{13} sin (5 t) - \frac{2}{13} cos (5 t)

y = c_{1} e^{- t} + c_{2} e^{- 5 t} + \frac{3}{13} sin (5 t) - \frac{2}{13} cos (5 t)

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (3 t)

\frac{d ^{2} y}{d x ^{2}} + 9 y = - 15 x + 7 + e^{x}

y^{^{''}} + 9 y = sin (3 x) + e^{2 x}

y^{^{''}} - y^{^{'}} - 2 y = 8 x + 5

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

What is the solution for y^{''}+6y^'+5y=10cos(5t) ?
The solution for y^{''}+6y^'+5y=10cos(5t) is y=c_{1}e^{-t}+c_{2}e^{-5t}+3/13 sin(5t)-2/13 cos(5t)