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

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

y^{^{''}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = 22

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)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 15 y = 0

y^{^{''}} - 19 y^{^{'}} - 10 y = 0, y (0) = 1, y^{^{'}} (0) = 1

x^{^{'''}} - 2 x^{^{''}} - x^{^{'}} + 2 x = 0

y^{^{''}} + 40 y^{^{'}} + 4000 y = 0

y^{^{''}} + 20 y^{^{'}} + 64 y = 0, y (0) = 5, y^{^{'}} (0) = 0

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