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

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

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

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

y = e^{t} - 1

Solution steps

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

Solve linear ODE:

y = e^{t} - 1

y = e^{t} - 1

2 y^{^{''}} + 14 y^{^{'}} + 12 y = 0

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 y = 0

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

y^{^{'''}} + 4 y^{^{''}} + y^{^{'}} - 26 y = 0, y (0) = e^{2 x}

y^{^{''}} - 2 y^{^{'}} = - 4 y

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