What is the solution for y^{''}+4y^'=2e^{-t},y(0)=2,y^'(0)=2 ?

The solution for y^{''}+4y^'=2e^{-t},y(0)=2,y^'(0)=2 is y= 1/3+8/3-1/3 e^{-4t}-2/3 e^{-t}

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

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

y = \frac{1}{3} + \frac{8}{3} - \frac{1}{3} e^{- 4 t} - \frac{2}{3} e^{- t}

Solution steps

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

Solve linear ODE:

y = \frac{1}{3} + \frac{8}{3} - \frac{1}{3} e^{- 4 t} - \frac{2}{3} e^{- t}

y = \frac{1}{3} + \frac{8}{3} - \frac{1}{3} e^{- 4 t} - \frac{2}{3} e^{- t}

(D^{2} - 2 D + 1) y = 4 x^{2} e^{x}

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

4 y^{^{''}} - 20 y^{^{'}} + 25 y = 6 - 5 x + 50 x^{2}

y^{^{''}} - 4 y^{^{'}} = sin^{2} (x)

y^{^{''}} - 4 y^{^{'}} + 5 y = 5 x^{2} + 2 x

What is the solution for y^{''}+4y^'=2e^{-t},y(0)=2,y^'(0)=2 ?
The solution for y^{''}+4y^'=2e^{-t},y(0)=2,y^'(0)=2 is y= 1/3+8/3-1/3 e^{-4t}-2/3 e^{-t}