What is the solution for y^{''}-2y^'+2y=0,y(pi/2)=0,y(pi)=-1 ?

The solution for y^{''}-2y^'+2y=0,y(pi/2)=0,y(pi)=-1 is y=e^{t-pi}cos(t)

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y^{''}-2y^'+2y=0,y(pi/2)=0,y(pi)=-1

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (\frac{π}{2}) = 0, y (π) = - 1

y = e^{t - π} cos (t)

Solution steps

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

Solve linear ODE:

y = e^{t - π} cos (t)

y = e^{t - π} cos (t)

15 y^{^{''}} - 4 y^{^{'}} - 3 y = 0

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

y^{^{'''}} - 8 y^{^{''}} + 16 y^{^{'}} = 0

y^{^{''}} + 5 y^{^{'}} - 6 y = 0, y (0) = - 5, y^{^{'}} (0) = 37

y^{^{''}} + 2 y^{^{'}} = - 4 y

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