What is the solution for x^{'''}-x^{''}=0,x(0)=1,x^'(0)=3,x^{''}(0)=2 ?

The solution for x^{'''}-x^{''}=0,x(0)=1,x^'(0)=3,x^{''}(0)=2 is x=t+2e^t-1

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x^{'''}-x^{''}=0,x(0)=1,x^'(0)=3,x^{''}(0)=2

x^{^{'''}} - x^{^{''}} = 0, x (0) = 1, x^{^{'}} (0) = 3, x^{^{''}} (0) = 2

x = t + 2 e^{t} - 1

Solution steps

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

Solve linear ODE:

x = t + 2 e^{t} - 1

x = t + 2 e^{t} - 1

y^{^{''}} - 10 y^{^{'}} + 34 y = 0, y (0) = 5, y^{^{'}} (0) = - 2

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

4 y^{^{''}} + 28 y = 0, y (0) = 3, y^{^{'}} (0) = 7

y^{^{'''}} + y = 0, y (0) = 0, y^{^{'}} (0) = 1, y^{^{''}} (0) = 2

(D^{2} + 2 D^{2} + D + 2) y = 0

What is the solution for x^{'''}-x^{''}=0,x(0)=1,x^'(0)=3,x^{''}(0)=2 ?
The solution for x^{'''}-x^{''}=0,x(0)=1,x^'(0)=3,x^{''}(0)=2 is x=t+2e^t-1