What is the solution for y^{''}+16y=0,y(pi^2)=-10,y^'(pi^2)=3 ?

The solution for y^{''}+16y=0,y(pi^2)=-10,y^'(pi^2)=3 is y=c_{1}cos(4t)+c_{2}sin(4t)

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y^{''}+16y=0,y(pi^2)=-10,y^'(pi^2)=3

y^{^{''}} + 16 y = 0, y (π^{2}) = - 10, y^{^{'}} (π^{2}) = 3

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)

4 y^{^{''}} - 12 y^{^{'}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = 5

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 2

49 y^{^{''}} + 14 y^{^{'}} + y = 0

4 y^{^{''}} - 7 y^{^{'}} + 4 y = 0

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

What is the solution for y^{''}+16y=0,y(pi^2)=-10,y^'(pi^2)=3 ?
The solution for y^{''}+16y=0,y(pi^2)=-10,y^'(pi^2)=3 is y=c_{1}cos(4t)+c_{2}sin(4t)