Given the differential equation:
y’’+y=0
the general solution can be written as:
y=c₁*cos(t) + c₂*sin(t)

With the given boundary conditions, we will find the c₁ and c₂ that satisfy the equation.

1=c₁*1 + c₂*0
-1= -c₁*0 + c₂*1

1=c₁
-1=c₂
 

b) y(0)=0, y(^p/₂)=0
y= c₁*cos(t) + c₂*sin(t)

0=c₁*cos(0) + c₂*sin(0)
0= c₁*cos(^p/₂) + c₂*sin(^p/₂)

0=c₁
0=c₂
 

c) y(0)=0, y(p)=0
y= c₁*cos(t) + c₂*sin(t)

0=c₁*1 + c₂*0
0=c₁*-1 + c₂*0

0=c₁
c₂ has an infinite number of solutions.
 

d) y(0)=0, y(p)=1
y= c₁*cos(t) + c₂*sin(t)

0=c₁*1 + c₂*0
1=c₁*-1 + c₂*0

0=c₁= -1
c₁ has no solution
 

e) y(^p/₄)=1, y’(^p/₄)= -2
y= c₁*cos(t) + c₂*sin(t)
y’= -c₁*sin(t) + c₂*cos(t)

1=c₁*Ö(2)/2 + c₂*Ö(2)/2
-2=-c₁*Ö(2)/2 + c₂*Ö(2)/2

Ö(2)= c₁ + c₂
-2*Ö(2)= -c₁ + c₂

-Ö(2)= 2*c₂

c₂= -Ö(2)/2
c₁= 3*Ö(2)/2
 

f) y(^p/₆)=-2, y(^p/₃)=2
y= c₁*cos(t) + c₂*sin(t)

-2=c₁*Ö(3)/2 + c₂*¹/₂
2=c₁*¹/₂ + c₂*Ö(3)/2

-4= c₁*Ö(3) + c₂
4= c₁ + c₂*Ö(3)

-4= c₁*Ö(3) + c₂
-4*Ö(3) = -c₁*Ö(3) - c₂ *3

-4-4*Ö(3)= -2 * c₂

c₂=2+2*Ö(3)
c₁= -2 -2*Ö(3)