What is the solution for y^{''}+16y=0,y^'(0)=-2,y(0)=2 ?

The solution for y^{''}+16y=0,y^'(0)=-2,y(0)=2 is y=2cos(4t)-1/2 sin(4t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+16y=0,y^'(0)=-2,y(0)=2

y^{^{''}} + 16 y = 0, y^{^{'}} (0) = - 2, y (0) = 2

y = 2 cos (4 t) - \frac{1}{2} sin (4 t)

Solution steps

y^{''} (t) + 16 y = 0

Solve linear ODE:

y = 2 cos (4 t) - \frac{1}{2} sin (4 t)

y = 2 cos (4 t) - \frac{1}{2} sin (4 t)

y^{^{''}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 5

8 y^{^{''}} + c y^{^{'}} + 128 y = 0, y (0) = 5, y^{^{'}} (0) = 8

y^{^{''}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 6

y^{^{''}} - \frac{4 y ^{^{'}}}{3} + \frac{97 y}{36} = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{31}{2}

What is the solution for y^{''}+16y=0,y^'(0)=-2,y(0)=2 ?
The solution for y^{''}+16y=0,y^'(0)=-2,y(0)=2 is y=2cos(4t)-1/2 sin(4t)