What is the solution for y^{''}+2y^'+2y=3,y(0)=1,y^'(0)=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+2y=3,y(0)=1,y^'(0)=0

y^{^{''}} + 2 y^{^{'}} + 2 y = 3, y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 2 y = 3

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{- x}

(4 d^{2} + 12 d + 9) y = 2022

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 e^{2 x}

y^{^{''''}} - y = 3 x + cos (x)

y^{^{''}} - 2 y^{^{'}} + 10 = 0

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