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

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

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

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

y = c_{1} + (- 1 - 2 c_{3}) e^{t} + c_{3} t e^{t} + 2 t

Solution steps

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

Solve linear ODE:

y = c_{1} + (- 1 - 2 c_{3}) e^{t} + c_{3} t e^{t} + 2 t

y = c_{1} + (- 1 - 2 c_{3}) e^{t} + c_{3} t e^{t} + 2 t

y^{^{''}} - 2 y^{^{'}} + 2 y = t cos (2 t)

y^{^{''}} - 2 y^{^{'}} + y = x^{\frac{- 1}{2}} \cdot e^{x}

y^{^{''}} + 2 y^{^{'}} + y = e^{-} x ln (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 7 x

y^{^{''}} - y = x^{2} e^{- x}

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