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Saladsamurai
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[SOLVED] Question about a Linear 1st Order DE
Find the general solution to \frac{dy}{dx}=5y
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 \mu(x)=e^{\int P(x)dx}[/tex]<br /> Now if I solve by separation:<br /> <br /> \frac{1}{5}\frac{dy}{y}=dx<br /> <br /> 1/5\ln y=x+C <----is there a preference of where the C goes (with x or y)?By Var of Parameters:<br /> <br /> since P=-5, \mu(x)=e^{-5x}<br /> <br /> e^{-5x}*y=C<br /> <br /> Now is 1/5\ln y=x+C equivalent to e^{-5x}*y=C?I mean they must be, but I just can't see it.
Homework Statement
Find the general solution to \frac{dy}{dx}=5y
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 \mu(x)=e^{\int P(x)dx}[/tex]<br /> Now if I solve by separation:<br /> <br /> \frac{1}{5}\frac{dy}{y}=dx<br /> <br /> 1/5\ln y=x+C <----is there a preference of where the C goes (with x or y)?By Var of Parameters:<br /> <br /> since P=-5, \mu(x)=e^{-5x}<br /> <br /> e^{-5x}*y=C<br /> <br /> Now is 1/5\ln y=x+C equivalent to e^{-5x}*y=C?I mean they must be, but I just can't see it.
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