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

The solution for y^{''}+y^'=0,y(pi/2)=2,y^'(pi/2)=1 is y=-e^{-t+pi/2}+3

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

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

y = - e^{- t + \frac{π}{2}} + 3

Solution steps

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

Solve linear ODE:

y = - e^{- t + \frac{π}{2}} + 3

y = - e^{- t + \frac{π}{2}} + 3

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

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} - 10 y = 0

x^{^{''}} = 2 x

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 1, y^{^{'}} (0) = 1

What is the solution for y^{''}+y^'=0,y(pi/2)=2,y^'(pi/2)=1 ?
The solution for y^{''}+y^'=0,y(pi/2)=2,y^'(pi/2)=1 is y=-e^{-t+pi/2}+3