What is the solution for y^{''}-2y^'+2y=0,y(pi)=7,y(0)=7 ?

The solution for y^{''}-2y^'+2y=0,y(pi)=7,y(0)=7 is y=e^t(-7/(e^pi)cos(t)+c_{2}sin(t))

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y^{''}-2y^'+2y=0,y(pi)=7,y(0)=7

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (π) = 7, y (0) = 7

y = e^{t} (- \frac{7}{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{7}{e ^{π}} cos (t) + c_{2} sin (t))

y = e^{t} (- \frac{7}{e ^{π}} cos (t) + c_{2} sin (t))

y^{^{''}} - 4 y^{^{'}} + 5 y = 0

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

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

y^{^{''}} - 0.1 y = 0

y^{^{''}} + y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 1

What is the solution for y^{''}-2y^'+2y=0,y(pi)=7,y(0)=7 ?
The solution for y^{''}-2y^'+2y=0,y(pi)=7,y(0)=7 is y=e^t(-7/(e^pi)cos(t)+c_{2}sin(t))