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))

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = 7

D^{2} (D + 3)^{2} y = 0

(d^{4} + i^{2}) y = 0

y^{^{''}} + 9 y = 0, y (0) = 10, y^{^{'}} (0) = 0

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 3 y = 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))