What is the solution for y^{''}+16y=sin(4t) ?

The solution for y^{''}+16y=sin(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)-(tcos(4t))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+16y=sin(4t)

y^{^{''}} + 16 y = sin (4 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t) - \frac{t cos ( 4 t )}{8}

Solution steps

y^{''} (t) + 16 y = sin (4 t)

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t) - \frac{t cos ( 4 t )}{8}

y = c_{1} cos (4 t) + c_{2} sin (4 t) - \frac{t cos ( 4 t )}{8}

x \to 0 + lim (\frac{2}{x ^{2}})

\int \frac{1}{x ^{2} + 2 x + 10} d x

\frac{d}{d x} ((x + 1) \cdot e^{a r c t a n (x)})

\frac{d}{d x} (- e^{- x^{2}})

\frac{d y}{d t} = 3 y + y^{4}

What is the solution for y^{''}+16y=sin(4t) ?
The solution for y^{''}+16y=sin(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)-(tcos(4t))/8