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

7 y^{^{''}} + y = 0

\int e^{4 x} \frac{1}{5} d x

\frac{d y}{d t} = (5 + sin (t)) \cdot (\frac{y}{5})

\int \frac{x + 11}{( x - 4 ) ( x + 1 )} d x

y = a x + b

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