What is the solution for y^{''}-2y^'+6y=0 ?

The solution for y^{''}-2y^'+6y=0 is y=e^t(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^{''}-2y^'+6y=0

y^{^{''}} - 2 y^{^{'}} + 6 y = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

\int 4 sin^{2} (x) cos^{3} (x) d x

\frac{d y}{d x} = tan (y)

\frac{d}{d x} (\frac{x - 1}{x ^{2}})

\int x \cdot (7 x^{2} + x) d x

\int 8^{2} - x^{2} d x

What is the solution for y^{''}-2y^'+6y=0 ?
The solution for y^{''}-2y^'+6y=0 is y=e^t(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))