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Could someone explain me this simplification?

enter image description here

I cannot understand exact reason why $c_1$ is before $\exp$ function without being in another one.

Screenshot presents end of solution of this differential equation: $$\frac{{\rm d}y}{{\rm d}x}=(x^2−2)(y−4)$$

I will be very grateful for the solution of it entirely with all the steps even in the form of a photo of a sheet of paper

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    $\begingroup$ $e^{c_{1}}$ is itself a constant i.e $e^{c_{1}} = c_{2}$. To solve the ODE, just separate and integrate $$\frac{dy}{dx} = (x^{2} - 2)(y - 4) \implies \frac{y'}{y - 4} = (x^{2} - 2) \implies \int \frac{y'}{y - 4} dx = \int (x^{2} - 2) dx$$ $\endgroup$
    – Matthew Cassell
    Commented Feb 18, 2023 at 1:40
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    $\begingroup$ If there are no partial derivatives involved, then you should remove the partial-differential-equations tag. $\endgroup$
    – Gerry Myerson
    Commented Feb 18, 2023 at 1:41
  • $\begingroup$ @MatthewCassell So I'm right this c before function isn't equal to c in exponentiation and naming them same is incorrect? $\endgroup$
    – xKRISTOFx
    Commented Feb 18, 2023 at 11:41
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    $\begingroup$ Yes, the $c_{1}$ before the exponential function is not the same $c_{1}$ that is in the exponential function. Naming them the same is not an issue though; we often relabel things in maths. There is nothing wrong in letting $e^{c_{1}} \to c_{1}$ provided it is clear to the reader. This is why we often, for clarity, use another constant name rather than relabelling i.e letting $e^{c_{1}} = c_{2}$ (especially in high school and undergraduate courses where there is a lot of potential for confusion). $\endgroup$
    – Matthew Cassell
    Commented Feb 18, 2023 at 12:27
  • $\begingroup$ Never mind, I fixed it myself. $\endgroup$
    – Gerry Myerson
    Commented Feb 19, 2023 at 11:59

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