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

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

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

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

y = e^{- t} cos (t)

Solution steps

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

Solve linear ODE:

y = e^{- t} cos (t)

y = e^{- t} cos (t)

x^{^{''}} + 6 x^{^{'}} + 9 x = 0, x (0) = a, x^{^{'}} (0) = b

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

9 (\frac{d ^{2} y}{d x ^{2}}) + 12 (\frac{d y}{d x}) + 4 y = 0

2 y^{^{''}} + 4 y^{^{'}} + k y = 0

\frac{d ^{2} y}{d x ^{2}} (4) + \frac{d y}{d x} (5) + y = 0

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