What is the solution for 2y^{''}-y=4sin(3t) ?

The solution for 2y^{''}-y=4sin(3t) is y=c_{1}e^{t/(sqrt(2))}+c_{2}e^{-t/(sqrt(2))}-4/19 sin(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}-y=4sin(3t)

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

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- \frac{t}{2}} - \frac{4}{19} sin (3 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- \frac{t}{2}} - \frac{4}{19} sin (3 t)

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- \frac{t}{2}} - \frac{4}{19} sin (3 t)

\frac{d}{d x} (\frac{3 x ^{2} - 1}{x 1 - x ^{2}})

f (x) = 9 x - 9 x

f (x) = - 3 cos (x), at x = \frac{- π}{3}

x \to 0 lim ((1 - x)^{- \frac{1}{x}})

\frac{d}{d x} (ln (\frac{x ^{2} - 3 x + 2}{x ^{2} + 1}))

What is the solution for 2y^{''}-y=4sin(3t) ?
The solution for 2y^{''}-y=4sin(3t) is y=c_{1}e^{t/(sqrt(2))}+c_{2}e^{-t/(sqrt(2))}-4/19 sin(3t)