What is the solution for y^{''}-2y^'+y=30e^tt^{-7} ?

The solution for y^{''}-2y^'+y=30e^tt^{-7} is y=c_{1}e^t+c_{2}te^t+(e^t)/(t^5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+y=30e^tt^{-7}

y^{^{''}} - 2 y^{^{'}} + y = 30 e^{t} t^{- 7}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{t ^{5}}

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = 30 e^{t} t^{- 7}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{t ^{5}}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{t ^{5}}

y^{^{''}} + 4 y = 12 t

y^{^{''}} - y^{^{'}} + 2 y = 2 e^{2 t} + 5 t - 1

y^{^{''}} + y^{^{'}} + y = k

4 y^{^{''}} - y = e^{\frac{1}{2} x} + 7

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 3 y = x^{2} e^{- 3 x}

What is the solution for y^{''}-2y^'+y=30e^tt^{-7} ?
The solution for y^{''}-2y^'+y=30e^tt^{-7} is y=c_{1}e^t+c_{2}te^t+(e^t)/(t^5)