What is the solution for y^{''}-4y^'+4y=t^{-5}e^{2t} ?

The solution for y^{''}-4y^'+4y=t^{-5}e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t})/(12t^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+4y=t^{-5}e^{2t}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 5} e^{2 t}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{12 t ^{3}}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = t^{- 5} e^{2 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{12 t ^{3}}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{12 t ^{3}}

x \to - 8 lim (x + 5)

\int \frac{1}{cos ^{2} ( 6 x )} d x

\frac{\partial}{\partial y} ((x^{2} + y) (x - y^{2}))

\int_{0}^{1} π (3 x)^{2} d x

\frac{d}{d x} (7^{x})

What is the solution for y^{''}-4y^'+4y=t^{-5}e^{2t} ?
The solution for y^{''}-4y^'+4y=t^{-5}e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t})/(12t^3)