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}

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