What is the solution for y^{''}+2y^'=0,y(0)=-28,y^'(0)=2 ?

The solution for y^{''}+2y^'=0,y(0)=-28,y^'(0)=2 is y=-e^{-2t}-27

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y^{''}+2y^'=0,y(0)=-28,y^'(0)=2

y^{^{''}} + 2 y^{^{'}} = 0, y (0) = - 28, y^{^{'}} (0) = 2

y = - e^{- 2 t} - 27

Solution steps

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

Solve linear ODE:

y = - e^{- 2 t} - 27

y = - e^{- 2 t} - 27

y^{^{''}} + 2 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} + 7 y = 0

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

y^{^{''}} + 3 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = 2

2 y^{^{''}} - 4 y^{^{'}} + 13 y = 0

What is the solution for y^{''}+2y^'=0,y(0)=-28,y^'(0)=2 ?
The solution for y^{''}+2y^'=0,y(0)=-28,y^'(0)=2 is y=-e^{-2t}-27