What is the solution for y^{''}+2y^'=3+4sin(2t) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'=3+4sin(2t)

y^{^{''}} + 2 y^{^{'}} = 3 + 4 sin (2 t)

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

Solution steps

y^{''} (t) + 2 y^{'} (t) = 3 + 4 sin (2 t)

Solve linear ODE:

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

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

\int_{1}^{22} 1 d x

\int 8 x^{2} cos (5 x) d x

f (x) = \frac{x}{x ^{2} + 10}

\int 10 sin (x) d x

x \to 2 + lim (\frac{x}{∣ x ∣})

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