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Saladsamurai
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[SOLVED] Question about a Linear 1st Order DE
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]<br /> Now if I solve by separation:<br /> <br /> [tex]\frac{1}{5}\frac{dy}{y}=dx[/tex]<br /> <br /> [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:<br /> <br /> since P=-5, [itex]\mu(x)=e^{-5x}[/itex]<br /> <br /> [tex]e^{-5x}*y=C[/tex]<br /> <br /> 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.[/itex]
Homework Statement
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]<br /> Now if I solve by separation:<br /> <br /> [tex]\frac{1}{5}\frac{dy}{y}=dx[/tex]<br /> <br /> [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:<br /> <br /> since P=-5, [itex]\mu(x)=e^{-5x}[/itex]<br /> <br /> [tex]e^{-5x}*y=C[/tex]<br /> <br /> 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.[/itex]
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