What is the solution for y^{''}+2y^'=0,y(0)=3,y^'(0)=10 ?

The solution for y^{''}+2y^'=0,y(0)=3,y^'(0)=10 is y=-5e^{-2t}+8

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y^{''}+2y^'=0,y(0)=3,y^'(0)=10

y^{^{''}} + 2 y^{^{'}} = 0, y (0) = 3, y^{^{'}} (0) = 10

y = - 5 e^{- 2 t} + 8

Solution steps

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

Solve linear ODE:

y = - 5 e^{- 2 t} + 8

y = - 5 e^{- 2 t} + 8

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

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

49 y^{^{''}} - 16 y = 0, y (0) = 1, y^{^{'}} (0) = 0

x^{^{''}} + 4 x^{^{'}} + 13 x = 0

y^{^{''}} - 5 y^{^{''}} + 4 y = 0

What is the solution for y^{''}+2y^'=0,y(0)=3,y^'(0)=10 ?
The solution for y^{''}+2y^'=0,y(0)=3,y^'(0)=10 is y=-5e^{-2t}+8