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

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

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

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

y = e^{t} (- \frac{1}{e ^{π}} cos (t) + c_{2} sin (t))

Solution steps

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

Solve linear ODE:

y = e^{t} (- \frac{1}{e ^{π}} cos (t) + c_{2} sin (t))

y = e^{t} (- \frac{1}{e ^{π}} cos (t) + c_{2} sin (t))

\frac{1}{8} y^{^{''}} + y^{^{'}} + 2 y = 0, y (0) = - 1, y^{^{'}} (0) = 12

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

x^{^{''}} + a x^{^{'}} = 0

y^{^{'''}} - y^{^{''}} - 16 y^{^{'}} - 20 y = 0

y^{^{''}} + 40 y^{^{'}} + 2000 y = 0, y (0) = 5, y^{^{'}} (0) = 0

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