What is the solution for y^{''}-2y^'+10y=0 ?

The solution for y^{''}-2y^'+10y=0 is y=e^t(c_{1}cos(3t)+c_{2}sin(3t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+10y=0

y^{^{''}} - 2 y^{^{'}} + 10 y = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (e^{\frac{4}{x}})

x \to 2 lim (5 x - 6)

\frac{d}{d x} (\frac{4 x - 8}{x - 5})

\int cot (4 t) 1 - sin^{2} (4 t) d t

\int \frac{5 x ^{2} - 5 x + 14}{( x - 2 ) ( x ^{2} + 4 )} d x

What is the solution for y^{''}-2y^'+10y=0 ?
The solution for y^{''}-2y^'+10y=0 is y=e^t(c_{1}cos(3t)+c_{2}sin(3t))