What is the solution for y^{''}-9y=4e^{3t},y(0)=0,y^'(0)=-1 ?

The solution for y^{''}-9y=4e^{3t},y(0)=0,y^'(0)=-1 is y=-5/18 e^{3t}+5/18 e^{-3t}+(2e^{3t}t)/3

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y^{''}-9y=4e^{3t},y(0)=0,y^'(0)=-1

y^{^{''}} - 9 y = 4 e^{3 t}, y (0) = 0, y^{^{'}} (0) = - 1

y = - \frac{5}{18} e^{3 t} + \frac{5}{18} e^{- 3 t} + \frac{2 e ^{3 t} t}{3}

Solution steps

y^{''} (t) - 9 y = 4 e^{3 t}

Solve linear ODE:

y = - \frac{5}{18} e^{3 t} + \frac{5}{18} e^{- 3 t} + \frac{2 e ^{3 t} t}{3}

y = - \frac{5}{18} e^{3 t} + \frac{5}{18} e^{- 3 t} + \frac{2 e ^{3 t} t}{3}

3 y^{^{''}} - 6 y^{^{'}} + 30 y = e^{x} \cdot tan (3 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 cos^{2} (4 x)

y^{^{''}} + 25 y^{^{'}} + 100 y = 90

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = 2 - 24 e^{x} + 40 e^{5 x}

y^{^{''}} - 16 y = e^{- 5 x} (- 8 - 9 x)

What is the solution for y^{''}-9y=4e^{3t},y(0)=0,y^'(0)=-1 ?
The solution for y^{''}-9y=4e^{3t},y(0)=0,y^'(0)=-1 is y=-5/18 e^{3t}+5/18 e^{-3t}+(2e^{3t}t)/3