What is the solution for y^{''}+4y=12sin(2t) ?

The solution for y^{''}+4y=12sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-3tcos(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=12sin(2t)

y^{^{''}} + 4 y = 12 sin (2 t)

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

Solution steps

y^{''} (t) + 4 y = 12 sin (2 t)

Solve linear ODE:

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

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

\int e^{3 x} (3 x + 1) d x

\int_{0}^{t} cos^{2} (wt) d t

\int 3 a y^{2} d y

\frac{d}{d x} (\frac{x + 5}{x - 1})

x \to 0 lim (sin (\frac{π x}{2}))

What is the solution for y^{''}+4y=12sin(2t) ?
The solution for y^{''}+4y=12sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-3tcos(2t)