What is the solution for y^{''}+y^'=0,y(pi/3)=16,y^'(pi/3)=-32 ?

The solution for y^{''}+y^'=0,y(pi/3)=16,y^'(pi/3)=-32 is y=32e^{-t+pi/3}-16

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y^{''}+y^'=0,y(pi/3)=16,y^'(pi/3)=-32

y^{^{''}} + y^{^{'}} = 0, y (\frac{π}{3}) = 16, y^{^{'}} (\frac{π}{3}) = - 32

y = 32 e^{- t + \frac{π}{3}} - 16

Solution steps

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

Solve linear ODE:

y = 32 e^{- t + \frac{π}{3}} - 16

y = 32 e^{- t + \frac{π}{3}} - 16

(10 d^{3} + d^{2} - 7 d + 2) y = 0

y^{^{'''}} + 3 y^{^{''}} - 8 y^{^{'}} - 30 y = 0

y^{^{'''}} - 7 y^{^{''}} - 8 y^{^{'}} + 16 y = 0

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

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{39}{2}

What is the solution for y^{''}+y^'=0,y(pi/3)=16,y^'(pi/3)=-32 ?
The solution for y^{''}+y^'=0,y(pi/3)=16,y^'(pi/3)=-32 is y=32e^{-t+pi/3}-16