What is the solution for y^{''}-y^'=1,y(0)=19,y^'(0)=1 ?

The solution for y^{''}-y^'=1,y(0)=19,y^'(0)=1 is y=2e^t+17-t

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y^{''}-y^'=1,y(0)=19,y^'(0)=1

y^{^{''}} - y^{^{'}} = 1, y (0) = 19, y^{^{'}} (0) = 1

y = 2 e^{t} + 17 - t

Solution steps

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

Solve linear ODE:

y = 2 e^{t} + 17 - t

y = 2 e^{t} + 17 - t

d, d (x) = \frac{5}{4} x^{2} + 56 x^{4} + \frac{8}{x} - 9 x

y^{^{''}} + 8 y^{^{''}} - 9 y = - 10 e^{- x} - 425 cos (2 x)

y^{^{''}} + 16 y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

9 y^{^{''}} - 6 y^{^{'}} + y = 50 sin (x)

4 y^{^{''}} + 4 y^{^{'}} + y = x e^{\cdot \frac{x}{2}} sin (x)

What is the solution for y^{''}-y^'=1,y(0)=19,y^'(0)=1 ?
The solution for y^{''}-y^'=1,y(0)=19,y^'(0)=1 is y=2e^t+17-t