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

The solution for y^{''}-2y^'+2y=0,y^'(pi)=0,y(1)=0 is y=0

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

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

y = 0

Solution steps

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

Solve linear ODE:

y = 0

y = 0

y^{^{''}} = 5 y, y (0) = - 2, y (2) = 10

y^{^{''}} + 9 y = 0, y^{1} (0) = sin (3 x)

y^{^{''}} + 16 y = 0, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + 16 y = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 8 y^{^{'}} + 32 y = 0, y (0) = 5, y^{^{'}} (0) = - 4

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