What is the solution for y^{''}+1y=0,y(pi^2)=3,y^'(pi^2)=-1 ?

The solution for y^{''}+1y=0,y(pi^2)=3,y^'(pi^2)=-1 is y=c_{1}cos(t)+c_{2}sin(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+1y=0,y(pi^2)=3,y^'(pi^2)=-1

y^{^{''}} + 1 y = 0, y (π^{2}) = 3, y^{^{'}} (π^{2}) = - 1

y = c_{1} cos (t) + c_{2} sin (t)

Solution steps

y^{''} (t) + 1 y = 0

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t)

y = c_{1} cos (t) + c_{2} sin (t)

- 2 y^{^{''}} + 4 y^{^{'}} = 0

8 x^{^{''}} + 20 x^{^{'}} + 32 x = 0

(D^{4} + 18 D^{2} + 81) y = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = 2, y^{^{'}} (0) = 0

2 y^{^{''}} + 8 y = 0, y^{^{'}} (0) = 1, y (0) = 0

What is the solution for y^{''}+1y=0,y(pi^2)=3,y^'(pi^2)=-1 ?
The solution for y^{''}+1y=0,y(pi^2)=3,y^'(pi^2)=-1 is y=c_{1}cos(t)+c_{2}sin(t)