What is the solution for y^{''}+4y=0,y^'(3pi)=4,y(0)=4 ?

The solution for y^{''}+4y=0,y^'(3pi)=4,y(0)=4 is y=4cos(2t)+2sin(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=0,y^'(3pi)=4,y(0)=4

y^{^{''}} + 4 y = 0, y^{^{'}} (3 π) = 4, y (0) = 4

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

Solution steps

y^{''} (t) + 4 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 8

10 y^{^{''}} + 11 y^{^{'}} + 3 y = 0

y^{^{'''}} - 5 y^{^{''}} + 3 y + 9 y = 0

9 y^{^{''}} - 64 y = 0

y^{^{''}} - 25 y^{^{'}} + 144 y = 0, y (0) = 2, y^{^{'}} (0) = 5

What is the solution for y^{''}+4y=0,y^'(3pi)=4,y(0)=4 ?
The solution for y^{''}+4y=0,y^'(3pi)=4,y(0)=4 is y=4cos(2t)+2sin(2t)