Respuesta :

JцstinBieber
[tex]\displaystyle\frac{dx}{dt} = 1 - t + x - tx\ \Rightarrow\ \frac{dx}{dt} = 1(1- t) + x(1 - t) \ \Rightarrow \\ \\ \frac{dx}{dt} = (1+x)(1 - t) \ \Rightarrow\ \int \frac{dx}{1+x} = \int (1 - t)\ dt\ \Rightarrow \\ \\ \textstyle \ln|1 + x| = t - \frac{1}{2}t^2 + C\ \Rightarrow\ |1 + x| = e^{t - t^2/2 + C }\ \Rightarrow \\ \\ 1 + x = \pm e^{t - t^2/2} \cdot e^C\Rightarrow \\ \\ x = -1 + Ke^{t - t^2/2},\ \text{where $K$ is any nonzero constant.}[/tex]

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