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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = \frac{x}{( 1 + x ^{2} )}

y = \frac{2}{3} x^{3} - \frac{3}{4 x ^{2}}

\frac{\partial}{\partial y} (\frac{cos ( y )}{x z ^{2}})

y^{^{'''}} = 6

\int sin (3 π t) d t

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