What is the solution for y^{''''}-y=0 ?

The solution for y^{''''}-y=0 is y=c_{1}e^t+c_{2}e^{-t}+c_{3}cos(t)+c_{4}sin(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''''}-y=0

y^{^{''''}} - y = 0

y = c_{1} e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t)

Solution steps

y^{(4)} (t) - y = 0

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t)

y = c_{1} e^{t} + c_{2} e^{- t} + c_{3} cos (t) + c_{4} sin (t)

\int x^{- 1} d x

\int_{0}^{2} e^{- t} d t

x \to \infty lim (8 + \frac{5}{x})

\frac{d}{d x} (- \frac{5}{( x - 2 ) ^{2}})

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

What is the solution for y^{''''}-y=0 ?
The solution for y^{''''}-y=0 is y=c_{1}e^t+c_{2}e^{-t}+c_{3}cos(t)+c_{4}sin(t)