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

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

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

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

y = - 4 e^{\frac{π}{2} - t} cos (t)

Solution steps

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

Solve linear ODE:

y = - 4 e^{\frac{π}{2} - t} cos (t)

y = - 4 e^{\frac{π}{2} - t} cos (t)

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