What is the solution for y^{''}+25y=0,y^'(pi)=5,y(0)=4 ?

The solution for y^{''}+25y=0,y^'(pi)=5,y(0)=4 is y=4cos(5t)-sin(5t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+25y=0,y^'(pi)=5,y(0)=4

y^{^{''}} + 25 y = 0, y^{^{'}} (π) = 5, y (0) = 4

y = 4 cos (5 t) - sin (5 t)

Solution steps

y^{''} (t) + 25 y = 0

Solve linear ODE:

y = 4 cos (5 t) - sin (5 t)

y = 4 cos (5 t) - sin (5 t)

y^{^{''}} + 64 y = 0, y (0) = 4, y^{^{'}} (0) = 1

y^{^{''}} + 16 y^{^{'}} + 3 y = 0

y^{^{''}} - 25 y^{^{'}} + 5 y = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (1) = 0, y^{^{'}} (1) = - 1

4 y^{^{''}} + 75 y = 0

What is the solution for y^{''}+25y=0,y^'(pi)=5,y(0)=4 ?
The solution for y^{''}+25y=0,y^'(pi)=5,y(0)=4 is y=4cos(5t)-sin(5t)