What is the solution for x^{''}+3x^'-4x=-4t+3 ?

The solution for x^{''}+3x^'-4x=-4t+3 is x=c_{1}e^t+c_{2}e^{-4t}+t

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x^{''}+3x^'-4x=-4t+3

x^{^{''}} + 3 x^{^{'}} - 4 x = - 4 t + 3

x = c_{1} e^{t} + c_{2} e^{- 4 t} + t

Solution steps

x^{''} (t) + 3 x^{'} (t) - 4 x = - 4 t + 3

Solve linear ODE:

x = c_{1} e^{t} + c_{2} e^{- 4 t} + t

x = c_{1} e^{t} + c_{2} e^{- 4 t} + t

y^{^{''}} - 4 y^{^{'}} + 4 = sin (x)

y^{^{''}} + 2 y^{^{'}} = 4, y (0) = 3, y^{^{'}} (0) = 5

y^{^{''}} + y = t sin (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = x^{2} - 1

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 e^{2} x

What is the solution for x^{''}+3x^'-4x=-4t+3 ?
The solution for x^{''}+3x^'-4x=-4t+3 is x=c_{1}e^t+c_{2}e^{-4t}+t