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

The solution for y^{''}-4*y^'-12*y=sin(2*t) is y=c_{1}e^{6t}+c_{2}e^{-2t}-1/20 sin(2t)+1/40 cos(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

y^{^{''}} - 4 \cdot y^{^{'}} - 12 \cdot y = sin (2 \cdot t)

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{1}{20} sin (2 t) + \frac{1}{40} cos (2 t)

Solution steps

y^{''} (t) - 4 \cdot y^{'} (t) - 12 \cdot y = sin (2 \cdot t)

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{1}{20} sin (2 t) + \frac{1}{40} cos (2 t)

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{1}{20} sin (2 t) + \frac{1}{40} cos (2 t)

\int_{t + 2}^{t + 4} 3 d x

y = (4 x + 1)^{2}, (0, 1)

f (- 8) = 6 (1 - x)

x \to 1 - lim (\frac{x - 2}{x - 1})

\frac{d}{d x} (ln (\frac{x - 1}{x + 2}))

What is the solution for y^{''}-4*y^'-12*y=sin(2*t) ?
The solution for y^{''}-4*y^'-12*y=sin(2*t) is y=c_{1}e^{6t}+c_{2}e^{-2t}-1/20 sin(2t)+1/40 cos(2t)