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

The solution for y^{''}-2y^'+2y=6 is y=e^t(c_{1}cos(t)+c_{2}sin(t))+3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to \infty lim (1 - \frac{1}{x})

\int e^{- s t} cos (a t) d t

\int \frac{2}{x 1 + 2 x} d x

\int \frac{1}{t ^{2} - 2 t + 37} d t

\int_{0}^{\frac{π}{4}} 7 sec^{2} (x) d x

What is the solution for y^{''}-2y^'+2y=6 ?
The solution for y^{''}-2y^'+2y=6 is y=e^t(c_{1}cos(t)+c_{2}sin(t))+3