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**[SOLVED] Question about a Linear 1st Order DE**

**1. The problem statement, all variables and given/known data**

Find the general solution to [tex]\frac{dy}{dx}=5y[/tex]

Now I know that this is separable. But it is in the exercise set that immediately follows "finding a general solution" in which they use variation of parameters. At least I think that is what it is called...when for y'+P(x)y=f(x) you find the integrating factor [itex]\mu(x)=e^{\int P(x)dx}[/tex]

Now if I solve by separation:

[tex]\frac{1}{5}\frac{dy}{y}=dx[/tex]

[tex]1/5\ln y=x+C[/tex] <----is there a preference of where the C goes (with x or y)?

By Var of Parameters:

since P=-5, [itex]\mu(x)=e^{-5x}[/itex]

[tex]e^{-5x}*y=C[/tex]

Now is [tex]1/5\ln y=x+C[/tex] equivalent to [tex]e^{-5x}*y=C[/tex]?

I mean they must be, but I just can't see it.

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