What is the solution for y^{''}+16y=0,y(pi^8)=-2,y^'(pi^8)=16 ?

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

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y^{''}+16y=0,y(pi^8)=-2,y^'(pi^8)=16

y^{^{''}} + 16 y = 0, y (π^{8}) = - 2, y^{^{'}} (π^{8}) = 16

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^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 3, y (0) = 2

y^{^{''}} - \frac{10}{3} y^{^{'}} + α y = 0

(y^{^{''}} + 4 y)^{4} = 0

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

\frac{d ^{2} y}{d x ^{2}} + 0.25 \frac{d y}{d x} + 2 y = 0

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