What is the solution for y^{''}-9y^'=5e^{-2t},y(0)=1,y^'(0)=2 ?

The solution for y^{''}-9y^'=5e^{-2t},y(0)=1,y^'(0)=2 is y=(-1/2+17/22)e^{9t}+1/2+5/22 e^{-2t}

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y^{''}-9y^'=5e^{-2t},y(0)=1,y^'(0)=2

y^{^{''}} - 9 y^{^{'}} = 5 e^{- 2 t}, y (0) = 1, y^{^{'}} (0) = 2

y = (- \frac{1}{2} + \frac{17}{22}) e^{9 t} + \frac{1}{2} + \frac{5}{22} e^{- 2 t}

Solution steps

y^{''} (t) - 9 y^{'} (t) = 5 e^{- 2 t}

Solve linear ODE:

y = (- \frac{1}{2} + \frac{17}{22}) e^{9 t} + \frac{1}{2} + \frac{5}{22} e^{- 2 t}

y = (- \frac{1}{2} + \frac{17}{22}) e^{9 t} + \frac{1}{2} + \frac{5}{22} e^{- 2 t}

y^{^{''}} - 4 y = 4 x e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + y = sin (3 x) cos (2 x)

3 y^{^{''}} + 3 y^{^{'}} - 1 = 0

5 y^{^{''}} + y^{^{'}} = - 6 x, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{'''}} - 2 y^{^{''}} + 2 y^{^{'}} = 1 + x e^{x} e

What is the solution for y^{''}-9y^'=5e^{-2t},y(0)=1,y^'(0)=2 ?
The solution for y^{''}-9y^'=5e^{-2t},y(0)=1,y^'(0)=2 is y=(-1/2+17/22)e^{9t}+1/2+5/22 e^{-2t}