What is the solution for y^{''}-10y^'+27y=0 ?

The solution for y^{''}-10y^'+27y=0 is y=e^{5t}(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^{''}-10y^'+27y=0

y^{^{''}} - 10 y^{^{'}} + 27 y = 0

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

Solution steps

y^{''} (t) - 10 y^{'} (t) + 27 y = 0

Solve linear ODE:

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

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

\int x^{2} 7 x + 6 d x

\frac{d}{d x} (x^{2} + x y + y^{2} + \frac{3}{x} + \frac{3}{y} + 5)

\int 11 sin (\frac{π t}{12}) + 10 d t

n = 0 \sum \infty 0. 5^{n}

y^{^{'}} - \frac{2}{x} y = x^{2} e^{x}

What is the solution for y^{''}-10y^'+27y=0 ?
The solution for y^{''}-10y^'+27y=0 is y=e^{5t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))