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

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (π) = e^{π}, y^{^{'}} (π) = 0

y = e^{t} (- cos (t) + sin (t))

Solution steps

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

Solve linear ODE:

y = e^{t} (- cos (t) + sin (t))

y = e^{t} (- cos (t) + sin (t))

y^{^{'''}} - 9 y^{^{''}} + 23 y^{^{'}} - 15 y = 0

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 4

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

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0

6 y^{^{''}} + y^{^{'}} - 5 y = 0

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