What is the solution for y^{''}-y^'=1,y(0)=18,y^'(0)=1 ?

The solution for y^{''}-y^'=1,y(0)=18,y^'(0)=1 is y=2e^t+16-t

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y^{''}-y^'=1,y(0)=18,y^'(0)=1

y^{^{''}} - y^{^{'}} = 1, y (0) = 18, y^{^{'}} (0) = 1

y = 2 e^{t} + 16 - t

Solution steps

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

Solve linear ODE:

y = 2 e^{t} + 16 - t

y = 2 e^{t} + 16 - t

y^{^{''}} + 5 y = (x)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = 12 e^{x}

y^{^{''}} - y^{^{'}} = 8 - (t - 2)

x^{^{''}} - 16 x^{^{'}} + 64 x = 5 t e^{8 t}

y^{^{''}} - 5 y^{^{'}} + 6 y = - 2 x + 1

What is the solution for y^{''}-y^'=1,y(0)=18,y^'(0)=1 ?
The solution for y^{''}-y^'=1,y(0)=18,y^'(0)=1 is y=2e^t+16-t