What is the solution for y^{''}+4y+4y=3,e^{-t}y(0)=2 ?

The solution for y^{''}+4y+4y=3,e^{-t}y(0)=2 is y=c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t)+3/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y+4y=3,e^{-t}y(0)=2

y^{^{''}} + 4 y + 4 y = 3, e^{- t} y (0) = 2

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t) + \frac{3}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t) + \frac{3}{8}

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t) + \frac{3}{8}

What is the solution for y^{''}+4y+4y=3,e^{-t}y(0)=2 ?
The solution for y^{''}+4y+4y=3,e^{-t}y(0)=2 is y=c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t)+3/8