What is the solution for y^{''}+16y=0,y(pi/2)=-4,y^'(pi/2)=8 ?

The solution for y^{''}+16y=0,y(pi/2)=-4,y^'(pi/2)=8 is y=-4cos(4t)+2sin(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(pi/2)=-4,y^'(pi/2)=8

y^{^{''}} + 16 y = 0, y (\frac{π}{2}) = - 4, y^{^{'}} (\frac{π}{2}) = 8

y = - 4 cos (4 t) + 2 sin (4 t)

Solution steps

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

Solve linear ODE:

y = - 4 cos (4 t) + 2 sin (4 t)

y = - 4 cos (4 t) + 2 sin (4 t)

36 y^{^{''}} - 108 y^{^{'}} + 81 y = 0, y (0) = 7, y^{^{'}} (0) = 1

y^{^{'''}} + 3 y^{^{''}} + y^{^{'}} - 205 y = 0

2 y^{^{''}} - 5 y^{^{'}} - y = 0

θ^{^{''}} (t) + \frac{9.81}{72} θ (t) = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = - 1, y^{^{'}} (0) = 12

What is the solution for y^{''}+16y=0,y(pi/2)=-4,y^'(pi/2)=8 ?
The solution for y^{''}+16y=0,y(pi/2)=-4,y^'(pi/2)=8 is y=-4cos(4t)+2sin(4t)