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

The solution for y^{''}-2y^'+y=te^t,y(0)=0,y^'(0)=0 is y=(e^tt^3)/6

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

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

y = \frac{e ^{t} t ^{3}}{6}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{t} t ^{3}}{6}

y = \frac{e ^{t} t ^{3}}{6}

y^{^{''}} - 9 y^{^{'}} + 14 y = 3 x^{2} - 5 sin (2 x) + 8 x e^{6 x}

2 y^{^{''}} + 4 y^{^{'}} + 2 y = \frac{1}{e ^{x} ( x + 1 )}

y^{^{''}} + y = x + 1

y^{^{''}} - 6 y + 5 y^{^{'}} = - 9 e^{2 x}

y^{^{'''}} + 21 y^{^{''}} - 1372 y = e^{- 14 x}

What is the solution for y^{''}-2y^'+y=te^t,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-2y^'+y=te^t,y(0)=0,y^'(0)=0 is y=(e^tt^3)/6