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

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

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

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

y = e^{t} - 1 - t

Solution steps

y^{''} (t) - y^{'} (t) = 1

Solve linear ODE:

y = e^{t} - 1 - t

y = e^{t} - 1 - t

y^{^{''}} + 9 y = cos (7 t)

(D^{2} + 3 D - 10) y = 20 + 5 e^{3 x}

(D^{4} + 4 D^{2}) y = sin (2 x) + x e^{x} + 4

y^{^{''}} + 5 \frac{d y}{d x} - 6 y = 2 + (- 2) sin (2 x)

y^{^{''}} - y^{^{'}} + 6 y = 3 sin (2 x) + 4 cos (2 x)

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