Solve the initial value problem

Question

 y″+y=0,   y(π3)=6, y′(π/3)=−12

Choices:

1.  y(t)=y(t)= 6cost−12sint
2. y(t)= √3 cost + (√3 + 6) sint
3. y(t)= −12cost + 6sint
4. y(t)= √3cost+√3sint
5. y(t)= cost+sint
6. y(t)= (3+6√3)cost+(3√3−6)sint
7. y(t)= e^t−e^−t
8. y(t)= c1cost+c2sint
9. y(t)= √3cost+6√3sint

Please include solution thank you!

Answer 1

Characteristic equation r² + 1 = 0 r = ±i.

y = Acosx + Bsinx;

y(π/3) = Acosπ/3 + Bsinπ/3 = 6, or A/2 + B√3/2 = 6; A + B√3 = 12

y' = - Asinx + Bcosx; y'(π/3) = - A√3/2 + B/2 = -12; A√3 - B = 24.

4A = 12 + 24√3; A = 3 + 6√3; B = 3√3 - 6

So, y = (3 + 6√3)cosx + (3√3 - 6)sinx