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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 5 lim (- 2 x)

f (x) = 3 x + \frac{2}{x}

x \to 0 lim (5 - \frac{x}{2})

x \to 8 - lim (\frac{x}{x - 8})

n = 1 \sum \infty 4 \cdot 3^{n}

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