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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} = 6 y

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - 4, y^{^{'}} (0) = 14

3 y^{^{''}} - 24 y^{^{'}} + 48 y = 0

y^{^{'''}} - 6 y^{^{''}} + 9 y = 0

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