What is the solution for y^{''}+16y=0,y(pi^8)=-4,y^'(pi^8)=-8 ?

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

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y^{''}+16y=0,y(pi^8)=-4,y^'(pi^8)=-8

y^{^{''}} + 16 y = 0, y (π^{8}) = - 4, 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^{^{''}} = - y, y (0) = 0

y^{^{''}} = 13 \cdot y

y^{^{''}} + 2 y^{^{'}} + 1 y = 0, y (0) = 6, y^{^{'}} (0) = - 1

y^{^{''}} - y^{^{'}} - 46 y^{^{'}} - 80 y = 0

y^{^{'}} + 4 y^{^{''}} - 5 y^{^{'}} = 0

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