What is the solution for y^{''}-2y^'+3y=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+3y=0

y^{^{''}} - 2 y^{^{'}} + 3 y = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 3 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{1}{3} sin^{3} (x) - sin (x) + 1)

x \to 4 lim (\frac{x + 4}{x + 7})

\frac{\partial}{\partial x} (e^{- 2 x} cos (6 y))

x \to x lim (25)

\int \frac{4 x ^{3}}{x ^{2} + 9} d x

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