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

The solution for y^{''}+2y^'+y=sin(2t) is y=c_{1}e^{-t}+c_{2}te^{-t}-3/25 sin(2t)-4/25 cos(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{- t} + c_{2} t e^{- t} - \frac{3}{25} sin (2 t) - \frac{4}{25} cos (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} - \frac{3}{25} sin (2 t) - \frac{4}{25} cos (2 t)

y = c_{1} e^{- t} + c_{2} t e^{- t} - \frac{3}{25} sin (2 t) - \frac{4}{25} cos (2 t)

\int_{- \infty}^{6} x e^{\frac{x}{2}} d x

y^{^{'}} = \frac{1}{1 + y}

x^{2} (1 - 3 x)

x \to 1 lim (- (3))

\frac{d}{d x} (ln (x) x)

What is the solution for y^{''}+2y^'+y=sin(2t) ?
The solution for y^{''}+2y^'+y=sin(2t) is y=c_{1}e^{-t}+c_{2}te^{-t}-3/25 sin(2t)-4/25 cos(2t)