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I have to solve the following equation

$$ z'-e^z=2$$

where $z=f(x)$, but I can't solve this. Could you please help me ?

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Try multiplying by $\mathrm{e}^{-z}$, as this will lead to $$ \mathrm{e}^{-z}z’ - 1 = 2\mathrm{e}^{-z} $$ Or $$ -(\mathrm{e}^{-z})’ - 1 = 2\mathrm{e}^{-z} $$ We can then re-write as $$ -y’ -1 = 2y $$ With $y=\mathrm{e}^{-z}$. Should be easier to solve.

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it looks like Bernouilli 's equation $$z'=e^z+2$$ substitute $$u=e^z+2 \implies \frac {du}{dz}=e^z=u-2$$ The equation becomes $$\frac {dz}{dt}=u$$ $$\frac {dz}{du}\frac {du}{dt}=u$$ $$\frac {du}{dt}=u\frac {du}{dz}$$ $$\frac {du}{dt}=u(u-2)$$ $$u'_t+2u=u^2$$ This last one is Bernouilli's equation. But it's also separable. And easier to integrate that way $$\int \frac {du}{u(u-2)}=\int dt$$

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