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

The solution for x^{''}+2x^'+x=e^{2t},x(0)=2,x^'(0)=1 is x= 17/9 e^{-t}+(8e^{-t}t)/3+1/9 e^{2t}

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

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

x = \frac{17}{9} e^{- t} + \frac{8 e ^{- t} t}{3} + \frac{1}{9} e^{2 t}

Solution steps

x^{''} (t) + 2 x^{'} (t) + x = e^{2 t}

Solve linear ODE:

x = \frac{17}{9} e^{- t} + \frac{8 e ^{- t} t}{3} + \frac{1}{9} e^{2 t}

x = \frac{17}{9} e^{- t} + \frac{8 e ^{- t} t}{3} + \frac{1}{9} e^{2 t}

What is the solution for x^{''}+2x^'+x=e^{2t},x(0)=2,x^'(0)=1 ?
The solution for x^{''}+2x^'+x=e^{2t},x(0)=2,x^'(0)=1 is x= 17/9 e^{-t}+(8e^{-t}t)/3+1/9 e^{2t}