What is the solution for y^{''}+16y=0,y(pi^8)=-3,y^'(pi^8)=8 ?

The solution for y^{''}+16y=0,y(pi^8)=-3,y^'(pi^8)=8 is y=c_{1}cos(4t)+c_{2}sin(4t)

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y^{''}+16y=0,y(pi^8)=-3,y^'(pi^8)=8

y^{^{''}} + 16 y = 0, y (π^{8}) = - 3, y^{^{'}} (π^{8}) = 8

y = c_{1} cos (4 t) + c_{2} sin (4 t)

Solution steps

y^{''} (t) + 16 y = 0

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t)

y^{^{''}} - \frac{1}{a ^{2}} y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

4 y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 1

y^{^{''}} + y = 0, y (\frac{π}{3}) = 2

9 y^{^{''}} + 24 y^{^{'}} - 33 y = 0

What is the solution for y^{''}+16y=0,y(pi^8)=-3,y^'(pi^8)=8 ?
The solution for y^{''}+16y=0,y(pi^8)=-3,y^'(pi^8)=8 is y=c_{1}cos(4t)+c_{2}sin(4t)