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This is one of my homework questions. I know it's not allowed but I can't seem to match my answer with the solution: You have to solve the initial value problem. $$xy' + y = 0,\quad y(4) = 6$$ My attempt: $$\frac{y'}{y} = \frac{-1}{x} = \int \frac{y'}{y} dx = \int \frac{-1}{x} dx = \int \frac{1}{y} dy = -\int \frac{1}{x} dx = ln(y) = -ln(x) + c$$

The solution they gave is $24/x$. I don't see how I can get to $24/x$ by substituting the initial values into that.

Where am I going wrong?

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Hint: $\ln y = c - \ln x$ is equivalent to $y = d/x$. –  Antonio Vargas Feb 15 at 18:51
    
An fyi: Our site policy is that homework questions are allowed here -- All we ask is that the asker shows their work, of which you have done an excellent job. :) –  anorton Feb 15 at 18:52
    
Thank you all. @Anorton, I assumed that the same no homework rule from some of the other stacks applied here. Sorry about that. –  Hisham Feb 15 at 19:09
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2 Answers

up vote 2 down vote accepted

Your solution $$\ln y+\ln x=c \\ \ln xy=c \\ y(4)=6 \implies c=\ln 24 \\ \ln xy=\ln 24 \\ xy=24$$

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I wasted so much time on this. I got up to c = ln 24 on my own and then didn't know how to get from that to 24/x :/ Thank you very much for this. –  Hisham Feb 15 at 19:02
    
Note that the function $\ln$ is one to one so $\ln a=\ln b\implies a=b$. Glad to help out. –  Semsem Feb 15 at 19:08
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From $\ln y=-\ln x +C$, we get, by exponentiating, $y=e^C x^{-1}$. Put $x=4$. Then $y=6$ and therefore $e^C=24$.

Detail: We have $e^{\ln y}=y$ and $e^{-\ln x}=(e^{\ln x})^{-1}=x^{-1}$.

Remark: It would have been easier to note that $x\frac{dy}{dx}+y$ is the derivative of $xy$. Thus $$\frac{d(xy)}{dx}=0,$$ and therefore $xy=D$ for some constant $D$. it follows that $y=\frac{D}{x}$.

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Thank you André. The only reason I marked Semsem's answer is because I found it easier to understand and closer to the steps I was taking. Thanks again. –  Hisham Feb 15 at 19:06
    
You are welcome. I myself prefer to "substitute immediately," before solving for $y$, so I prefer Semsem's path to my $e^C$ path. I have found, however, that students tend to prefer the $E^C$ path. You have good taste! –  André Nicolas Feb 15 at 19:09
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