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

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

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

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

y = e^{t} t^{2} + e^{t}

Solution steps

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

Solve linear ODE:

y = e^{t} t^{2} + e^{t}

y = e^{t} t^{2} + e^{t}

x^{^{''}} - x^{^{'}} - 6 x = 5 t

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 7 e x

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{- t}

y^{^{''}} - y^{^{'}} + 4 y = x^{3} e^{2 x} + x e^{2 x}

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

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