What is the solution for y^{''''}+4y^{''}=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''''}+4y^{''}=0

y^{^{''''}} + 4 y^{^{''}} = 0

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

Solution steps

y^{(4)} (t) + 4 y^{''} (t) = 0

Solve linear ODE:

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

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

x \to 1 lim (x ln (x) - x)

\frac{\partial}{\partial x} ((x + y) e^{y})

\int \frac{k}{r} (- \frac{r ^{2}}{4}) d r

\int \frac{1}{x ^{2} 400 - x ^{2}} d x

\frac{d}{d x} ((2 x^{3} + 4) (x^{4} - 4 x))

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