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

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

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

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

y = e^{t} - 1

Solution steps

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

Solve linear ODE:

y = e^{t} - 1

y = e^{t} - 1

y^{^{''}} + 8 y^{^{'}} + \frac{1}{2} y = 0

x^{^{''}} - 7 x^{^{'}} + 4 x = 0

x^{^{''}} + \frac{32}{2} x = 0, x (0) = - 1, x^{^{'}} (0) = - 6

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

9 y^{^{''}} + 6 y^{^{'}} + 82 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

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