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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+(y^')+y=0

y^{^{''}} + (y^{^{'}}) + y = 0

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

Solution steps

y^{''} (t) + (y^{'} (t)) + y = 0

Solve linear ODE:

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

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

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 2

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

2 y^{^{''}} + 8 y + 10 y = 0, y (0) = 5, y^{^{'}} (0) = 0

5 x^{^{''}} + 10 x^{^{'}} + 25 x = 0

u^{^{''}} + 7 u^{^{'}} + 2 H (t) = 0

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