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

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

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

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

y = - e^{2 t} + 1

Solution steps

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

Solve linear ODE:

y = - e^{2 t} + 1

y = - e^{2 t} + 1

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = - 2, y^{^{'}} (0) = 9

x^{^{''}} + 0.5 \cdot x^{^{'}} + \frac{17}{16} \cdot x = 0

3 x^{^{''}} + 2 x^{^{'}} + 3 x = 0

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = - 4

y^{^{''}} + 5 y^{^{'}} = 0, y (0) = 5, y^{^{'}} (0) = - 20

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