What is the solution for y^{''}-2y^'+8y=0 ?

The solution for y^{''}-2y^'+8y=0 is y=e^t(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-2y^'+8y=0

y^{^{''}} - 2 y^{^{'}} + 8 y = 0

y = e^{t} (c_{1} cos (7 t) + c_{2} sin (7 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = e^{t} (c_{1} cos (7 t) + c_{2} sin (7 t))

y = e^{t} (c_{1} cos (7 t) + c_{2} sin (7 t))

a re a - x^{2} + 3, 3 x^{2} - 9 x

\int_{0}^{a} x^{2} a^{2} - x^{2} d x

\int (3 x^{2} - 4 x + 5) d x

(5 cos (t))^{^{'}}

d er i v a t i v e y = \frac{6}{x ^{4}}

What is the solution for y^{''}-2y^'+8y=0 ?
The solution for y^{''}-2y^'+8y=0 is y=e^t(c_{1}cos(sqrt(7)t)+c_{2}sin(sqrt(7)t))