What is the solution for f^{''}(t)+9*f(t)=(-2*e^{(3t)}) ?

The solution for f^{''}(t)+9*f(t)=(-2*e^{(3t)}) is f(t)=c_{1}cos(3t)+c_{2}sin(3t)-1/9 e^{3t}

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f^{''}(t)+9f(t)=(-2e^{(3t)})

f^{^{''}} (t) + 9 \cdot f (t) = (- 2 \cdot e^{(3 t)})

f (t) = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{1}{9} e^{3 t}

Solution steps

f^{''} (t) + 9 \cdot f (t) = (- 2 \cdot e^{(3 t)})

Solve linear ODE:

f (t) = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{1}{9} e^{3 t}

f (t) = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{1}{9} e^{3 t}

y^{^{''}} + 4 y^{^{'}} + 5 y = 10 e^{- 3 x}

y^{^{''}} + 7 y^{^{'}} + 12 y = - 432 t^{3}

y^{^{''}} + v y^{^{'}} = c e^{(- r^{2})}

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- 4 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 e^{- 5 t}

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