What is the solution for y^{''}-4y^'+9y=0 ?

The solution for y^{''}-4y^'+9y=0 is y=e^{2t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+9y=0

y^{^{''}} - 4 y^{^{'}} + 9 y = 0

y = e^{2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y = e^{2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

\int_{0}^{2} x^{2} x^{3} + 1 d x

f (x) = \frac{7}{4 - x}, at x = - 1

\frac{\partial}{\partial x} (y^{4} sin (2 x))

x \to - 4 lim (\frac{x ^{2} - 9}{4 - x})

f (x) = 6 e^{- \frac{1}{6} x^{2}}

What is the solution for y^{''}-4y^'+9y=0 ?
The solution for y^{''}-4y^'+9y=0 is y=e^{2t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))