What is the solution for y^{''}+2y^'+10y=0,y(pi/2)=0,y^'(pi/2)=6 ?

The solution for y^{''}+2y^'+10y=0,y(pi/2)=0,y^'(pi/2)=6 is y=2e^{pi/2-t}cos(3t)

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y^{''}+2y^'+10y=0,y(pi/2)=0,y^'(pi/2)=6

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 6

y = 2 e^{\frac{π}{2} - t} cos (3 t)

Solution steps

y^{''} (t) + 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = 2 e^{\frac{π}{2} - t} cos (3 t)

y = 2 e^{\frac{π}{2} - t} cos (3 t)

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

0.093 y^{^{''}} + 6 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 0

2 y^{^{''}} + 3 y^{^{'}} = 0

2 y^{^{''}} - 7 y^{^{'}} + 3 y = 0

What is the solution for y^{''}+2y^'+10y=0,y(pi/2)=0,y^'(pi/2)=6 ?
The solution for y^{''}+2y^'+10y=0,y(pi/2)=0,y^'(pi/2)=6 is y=2e^{pi/2-t}cos(3t)