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

The solution for y^{''}-2y^'+y=te^t is y=c_{1}e^t+c_{2}te^t+(e^tt^3)/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-2y^'+y=te^t

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

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{3}}{6}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{3}}{6}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{3}}{6}

What is the solution for y^{''}-2y^'+y=te^t ?
The solution for y^{''}-2y^'+y=te^t is y=c_{1}e^t+c_{2}te^t+(e^tt^3)/6