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

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

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

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

y = - 7 e^{\frac{π}{2} - t} cos (t)

Solution steps

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

Solve linear ODE:

y = - 7 e^{\frac{π}{2} - t} cos (t)

y = - 7 e^{\frac{π}{2} - t} cos (t)

y^{^{''}} - 6 y^{^{'}} + 30 y = 0

33 y^{^{''''}} + 188 y^{^{''}} + 64 y = 0

x^{^{'''}} + 4 x^{^{''}} + x^{^{'}} - 6 x = 0

y^{^{''}} + 4 y = 0, y (\frac{π}{8}) = 0, y (\frac{π}{6}) = 1

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

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