What is the solution for (dx)/(dt)=1+t-x-tx ?

The solution for (dx)/(dt)=1+t-x-tx is x=-e^{-(t^2)/2-t-c_{1}}+1

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(dx)/(dt)=1+t-x-tx

\frac{d x}{d t} = 1 + t - x - t x

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

Solution steps

\frac{d x}{d t} = 1 + t - x - t x

Solve separable ODE:

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

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

\int 2 (2 x + 4)^{5} d x

\frac{d}{d x} (4^{4})

\int x^{2} cos (\frac{1}{5} x) d x

f (x) = (4 - x) (\frac{2}{x ^{2}} - 3)

y = \frac{28}{x}

What is the solution for (dx)/(dt)=1+t-x-tx ?
The solution for (dx)/(dt)=1+t-x-tx is x=-e^{-(t^2)/2-t-c_{1}}+1