What is the solution for y^{''}+4y^'+5y=0,y(pi/2)=0,y^'(pi/2)=5 ?

The solution for y^{''}+4y^'+5y=0,y(pi/2)=0,y^'(pi/2)=5 is y=-5e^{pi-2t}cos(t)

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y^{''}+4y^'+5y=0,y(pi/2)=0,y^'(pi/2)=5

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 5

y = - 5 e^{π - 2 t} cos (t)

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = - 5 e^{π - 2 t} cos (t)

y = - 5 e^{π - 2 t} cos (t)

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y^{^{'}} (0) = 2

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

y^{^{'''}} + 4 y^{^{''}} - 31 y^{^{'}} - 70 y = 0

y^{^{''}} - 18 y^{^{'}} + 81 y = 0, y (0) = 0, y^{^{'}} (0) = 8

4 y^{^{''}} + 8 y^{^{'}} = 0

What is the solution for y^{''}+4y^'+5y=0,y(pi/2)=0,y^'(pi/2)=5 ?
The solution for y^{''}+4y^'+5y=0,y(pi/2)=0,y^'(pi/2)=5 is y=-5e^{pi-2t}cos(t)