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

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

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

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

y = \frac{2}{9} e^{t} - \frac{2}{9} e^{- 2 t} + \frac{e ^{t} t}{3}

Solution steps

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

Solve linear ODE:

y = \frac{2}{9} e^{t} - \frac{2}{9} e^{- 2 t} + \frac{e ^{t} t}{3}

y = \frac{2}{9} e^{t} - \frac{2}{9} e^{- 2 t} + \frac{e ^{t} t}{3}

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{1}{x}

x^{^{''}} + 4 x^{^{'}} + 4 x = 100 sin (4 t)

x^{^{''}} + 50 x^{^{'}} + 400 x = 100 cos (20 t)

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

y^{^{''}} + 22 y^{^{'}} + 121 y = 16.3 u

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