# separation of variables Solve and prove

Using separation of variables Solve $$\frac{dy}{dx} = 3y +1.$$ My final answer I got was: $$y = \pm C \frac{e^x}{3} - \frac{1}{3}.$$

But I don't how to take the derivative of that to get it back to my orig problem

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"But I don't how to take the derivative of that" - What's the derivative of $e^x$? A constant? –  Ｊ. Ｍ. Oct 5 '11 at 16:24
But I don't how to take the derivative of that, to get it back to 3y +1. If you integrate something the derivative of that integral is the problem you started with yes? Well that is what I am trying to do (1/3)e^x != 3y+1 –  domshyra Oct 5 '11 at 16:36
There is a mistake in your integration, which probably came when you integrated $dy/(3y+1)$. The result should have been $\frac{1}{3} \ln(|3y+1|)$ (plus a constant). I think you left out the $1/3$ in front. So the solution you give to the DE is not right. It should involve $e^{3t}$. –  André Nicolas Oct 5 '11 at 16:37
Oh right!!! because I would use U-sub and yeah. Thank you!!! –  domshyra Oct 5 '11 at 16:48
Now it will pass the differentiation test! Differentiation for most people is (after a while) easy and accurate. So checking whether a claimed solution works is usually a good idea. –  André Nicolas Oct 5 '11 at 16:52

First, note that a constant divided by $3$ is just a constant, so you could rewrite what you have without the fraction. Also, plus or minus a constant is just a constant. So you could rewrite your solution as $$y = Ce^x - \frac{1}{3}.$$

To check, you just take the derivative: $C$ is a constant, so $$\frac{dy}{dx} = \frac{d}{dx}\left(Ce^x - \frac{1}{3}\right) = C\frac{d}{dx}e^x - \frac{d}{dx}\frac{1}{3} = Ce^x.$$

Then you compute $3y + 1 = 3Ce^x - 1 + 1 = 3Ce^x$.

This is not the same as $\frac{dy}{dx}$, so this is not a correct answer.

And, no, it wasn't my "simplifications" that did you in: $$\frac{d}{dx}\left( \pm \frac{Ce^x}{3} - \frac{1}{3}\right) = \pm\frac{C}{3}\frac{d}{dx}e^x = \pm\frac{C}{3}e^x,$$ and $$3\left(\pm\frac{Ce^x}{3}-\frac{1}{3}\right) + 1 = \pm Ce^x\neq \pm\frac{Ce^x}{3}.$$

So in fact, you did a mistake in your derivation.

From $$\frac{dy}{dx} = 3y+1,$$ we must consider the possibility that $3y+1=0$; this will occur if $y=-\frac{1}{3}$; that is one possible solution.

If $3y+1\neq 0$, then separating variables we get $$\frac{dy}{3y+1} = dx,$$ and integrating both sides, we have $$\int\frac{dy}{3y+1} = \int\,dx.$$ To do the integral on the left, we let $u=3y+1$; then $du = 3dy$, so $\frac{1}{3}du = dy$. Hence $$\int\frac{dy}{3y+1} = \int\frac{\frac{1}{3}du}{u} = \frac{1}{3}\int\frac{du}{u} = \frac{1}{3}\ln|u|+C = \frac{1}{3}\ln|3y+1|+C;$$ so, consolidating constants, we have: \begin{align*} \int\frac{dy}{3y+1} &= \int\,dx\\ \frac{1}{3}\ln|3y+1| &= x+C&&C\text{ an arbitrary constant}\\ \ln|3y+1| &= 3x + D&&D\text{ an arbitrary constant}\\ e^{\ln|3y+1|} &= e^{3x+D}\\ |3y+1| &= e^De^{3x}\\ |3y+1| &= Ae^{3x} &&A\text{ a positive constant}\\ 3y+1 &=\pm Ae^{3x} &&A\text{ a positive constant}\\ 3y+1 &= Be^{3x} &&B\text{ a non-zero constant}\\ 3y &= Be^{3x} - 1\\ y &= \frac{B}{3}e^{3x} - \frac{1}{3}\\ y &= Ke^{3x} - \frac{1}{3} &&K\text{ a non-zero constant} \end{align*} I set $D=3C$, still a constant, then $A=e^D$, a positive constant (because $e^D$ is always positive); then $B=\pm A$, a nonzero constant (because $A$ is never zero, so $B$ is never zero); and finally $K=\frac{B}{3}$, a nonzero constant (because $B$ is never zero).

So putting it together with the special case considered before, we have that the solutions are: $$\text{Either }y=-\frac{1}{3}\quad\text{or}\quad y = Ke^{3x}-\frac{1}{3}\text{ with }K\text{ non-zero.}$$ Then we may realize that if we allow $K=0$, then we get the special case, so we can summarize the two cases by saying $$y = Ke^{3x} - \frac{1}{3},\quad K\text{ an arbitrary constant.}$$

Now take derivatives and plug in to verify this is correct.

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wow you went all out in this explanation, and It was very helpful. Thank you very much! I really, really appreciate it! Thanks again!!! –  domshyra Oct 5 '11 at 16:59