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

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

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

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

y = - e^{- t} + 3

Solution steps

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

Solve linear ODE:

y = - e^{- t} + 3

y = - e^{- t} + 3

x \cdot \frac{d y}{d x} = y^{2} - y

(x^{2} + 3 x - 1) e^{x}

\frac{\partial}{\partial y} (y cos (x) + 2 x e^{y} - x)

f (x) = 3 x^{2} - 4 x + 6, at x = 2

f (x) = x^{3} - 3 x + 4, (3, 22)

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