Arbitrary constant C, when to resolve it to C?

Question

in the picture on top, we resolve the arbitrary constant from $-2C_1$ to C, by why don't we do so when it is $\frac{10}{C}$ (below)? In essence, why isn't $y=\frac {10}{e^x}+C$ correct?

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Answer 1

Yes, that $C$ is also an arbitrary constant. And therefore you can write it as, say, $10K$, in which case $\frac{10}y=e^z+C$ becomes $\frac{10}y=e^z+10K$, and therefore$$y=\frac1{e^z/10+K}.$$But you will never get $\frac{10}{e^z}+C$ by this process.

Answer 2

They are very different functions. Try to sketch them. As $x\rightarrow\infty$, one tends to $C$ and the other tends to $10/C$.

This basically comes down to the fact that you can’t write

$$\frac{b}{a+c}=\frac{b}{a}+\frac{b}{c}.$$

Sure, if, on the other hand, you had

$$\frac{e^x+C}{10}=\frac{e^x}{10}+\underbrace{\frac{C}{10}}_{C_1}$$

you could have relabelled the constants.

Answer 3

Look at the results. The solution space of a differential equation is not always of the form "some function plus a constant", especially if a $y$ term appears on the right-hand side. In other words, $$ \frac{\partial^2 (y(x) + C)}{\partial x^2} = \frac{\partial^2 y(x)}{\partial x^2} $$ but $$ -\frac{e^x (y(x)+C)^2}{10} \neq -\frac{e^x y(x)^2}{10} $$ (Maybe it's true for some values of $C$ such as $C=0$, but I'm writing that this is not an equality for all $C$.)

So the reasoning is different. So far, this has probably not been very helpful to you though. The thing that you need to remember when you're computing is that you can only replace $kC$ by $C$ if you're looking at the final solution, i.e. $y(x) = f(x) + kC$; in this case it is fine to replace $kC$ by $C$ since $C$ is arbitrary and the family $\{f(x) + kC \, | \, C \in \mathbb R \}$ is the same family as the family $\{f(x) + C \, | \, C \in \mathbb R\}$ because those are just two parametrizations of the same family.

Similarly, if $y(x) = e^C f(x)$ for some constant $C$ (as a solution of some other differential equation), you could re-write this as $y(x) = C f(x)$ for $C > 0$ since the families $\{e^Cf(x) \, | \, C \in \mathbb R \}$ and $\{Cf(x) \, | \, C > 0 \}$ parametrize the same solutions. So this statement about $C$ is really about re-parametrizing the solution space to simplify the description of the family, so when you do it, you need to check if you are describing the same space of solutions.

Hope that helps,