What is the solution for 3y^{''}+12y=-4sin(3t) ?

The solution for 3y^{''}+12y=-4sin(3t) is y=c_{1}cos(2t)+c_{2}sin(2t)+4/15 sin(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3y^{''}+12y=-4sin(3t)

3 y^{^{''}} + 12 y = - 4 sin (3 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{4}{15} sin (3 t)

Solution steps

3 y^{''} (t) + 12 y = - 4 sin (3 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{4}{15} sin (3 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{4}{15} sin (3 t)

y^{^{'''}} + 4 y^{^{'}} = 3 e^{x} (2 x + 5)

y^{^{''''}} - 4 y^{^{''}} = 5 x^{2} - e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 8} e^{2 t}

\frac{d ^{2} x}{d t ^{2}} + k \frac{d x}{d t} + a = 0

3 \frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 30 y = 15 sin (x)

What is the solution for 3y^{''}+12y=-4sin(3t) ?
The solution for 3y^{''}+12y=-4sin(3t) is y=c_{1}cos(2t)+c_{2}sin(2t)+4/15 sin(3t)