What is the solution for y^{''}+2y^'=8sin(2t)+32cos(2t) ?

The solution for y^{''}+2y^'=8sin(2t)+32cos(2t) is y=c_{1}+c_{2}e^{-2t}+3sin(2t)-5cos(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'=8sin(2t)+32cos(2t)

y^{^{''}} + 2 y^{^{'}} = 8 sin (2 t) + 32 cos (2 t)

y = c_{1} + c_{2} e^{- 2 t} + 3 sin (2 t) - 5 cos (2 t)

Solution steps

y^{''} (t) + 2 y^{'} (t) = 8 sin (2 t) + 32 cos (2 t)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 t} + 3 sin (2 t) - 5 cos (2 t)

y = c_{1} + c_{2} e^{- 2 t} + 3 sin (2 t) - 5 cos (2 t)

f (x) = 5 x^{2} - 6

f (x) = 4 x^{2} + 2 x, (- 1, 2)

- \frac{1}{2 x ^{2}} + \frac{32}{5 x ^{5}}

\int \frac{1}{x ^{- \frac{1}{2}}} d x

\frac{d}{d x} (x (4 x - 12)^{3})

What is the solution for y^{''}+2y^'=8sin(2t)+32cos(2t) ?
The solution for y^{''}+2y^'=8sin(2t)+32cos(2t) is y=c_{1}+c_{2}e^{-2t}+3sin(2t)-5cos(2t)