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

The solution for y^{''}+2y=0,y(0)=2,y^'(0)=2 is y=2cos(sqrt(2)t)+sqrt(2)sin(sqrt(2)t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = 2 cos (2 t) + 2 sin (2 t)

Solution steps

y^{''} (t) + 2 y = 0

Solve linear ODE:

y = 2 cos (2 t) + 2 sin (2 t)

y = 2 cos (2 t) + 2 sin (2 t)

7 y^{^{''''}} - 338 y^{^{''}} + 225 y = 0

\frac{d ^{2} y}{d x ^{2}} - 16 y = 0

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = 1

2 x^{^{''}} + 3 x^{^{'}} + 7 x = 0

y^{^{''}} = 3 y^{^{'}} + 5 y

What is the solution for y^{''}+2y=0,y(0)=2,y^'(0)=2 ?
The solution for y^{''}+2y=0,y(0)=2,y^'(0)=2 is y=2cos(sqrt(2)t)+sqrt(2)sin(sqrt(2)t)