What is the solution for y^{''}+2y^'+2y=sin(1t) ?

The solution for y^{''}+2y^'+2y=sin(1t) is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))+1/5 sin(t)-2/5 cos(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+2y=sin(1t)

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (1 t)

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 2 y = sin (1 t)

Solve linear ODE:

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

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

x \to 5 lim (- 2)

\frac{6}{( s - 2 ) ^{4}}

f (x) = x^{2} + 8

\int 4 x^{\frac{1}{3}} - 4 x^{- \frac{1}{3}} d x

\int - (sin (x))^{2} d x

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