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

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

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y^{''}-2y^'+y=3e^t,y(0)=1,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} + y = 3 e^{t}, y (0) = 1, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = 3 e^{t}

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} - 5 y = 125

y^{^{''}} + 4 y^{^{'}} + 5 y = - 5 x + 3 e^{- x}

y^{^{''}} + 4 y = 4

y^{^{''}} + y^{^{'}} - 12 y = 2 cos (t), y (0) = 1, y^{^{'}} (0) = 1

y^{^{'''}} + 8 y = 2 x - 5 + 8 e^{- 2 x}

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