What is the solution for y^{''}+16y=0,y^'(pi/4)=-3,y^'(0)=2 ?

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

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y^{''}+16y=0,y^'(pi/4)=-3,y^'(0)=2

y^{^{''}} + 16 y = 0, y^{^{'}} (\frac{π}{4}) = - 3, y^{^{'}} (0) = 2

y = c_{1} cos (4 t) + \frac{3}{4} sin (4 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (4 t) + \frac{3}{4} sin (4 t)

y = c_{1} cos (4 t) + \frac{3}{4} sin (4 t)

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 1, y^{^{'}} (0) = 0

y^{^{''}} - 3 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 5

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

y^{^{''}} + 25 y^{^{'}} + 2500 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} + 16 y = 0

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