8
$\begingroup$

Suppose we solve $\frac{dy}{dx} = \frac{1 + y}{2 + x} .$ Which can be written as the following and integrating both sides w.r.t. $y$ and $x$: $\int\frac{1}{1 + y}dy = \int\frac{1}{2 +x}dx ,$ we get $\ln(1+y) = \ln(2+x) + C$ One of the book says:

It's convenient to write the constant $C$ as the logarithm of some other constant $A$: $ \ln(1+y) = \ln(2+x) + \ln(A) \implies \ln A(2 + x)$ $ \therefore (1 + y) = A(2 + x)$

Question: Why is it "convenient to write the constant $C$ as the logarithm of some other constant $A$"? What liberty do we have to write $\ln(A)$ instead of just $C$? I think I am unaware of what a logarithm of a constant is. I mean the significance of it.

  • 0
    ${\int} \dfrac{1}{y}dy = \log|x| + c$2011-12-27

2 Answers 2

6

It is convenient because it allows you to write both sides of an equation as a logarithm, and then use the fact that $\ln(a)=\ln(b)$ implies $a=b$.

You have liberty to do so because $\ln$ is a surjective function, meaning that every real number has the form $\ln(A)$ for some $A>0$.

However, it is also unnecessary to do so, so do not worry if this style of argument seems unnatural at present. From $\ln(1+y)=\ln(2+x)+C$, you can exponentiate to get $1+y=e^C(2+x)$, and then recognize that $e^C$ is also an arbitrary (but positive) constant, and you might as well call it $A$.

A side note: $\ln(1+y)=\ln(2+x)+C$ is not quite a valid deduction, unless you are assuming that $y>-1$ and $x>-2$.

  • 1
    Usually you will be solving the DE given some *initial condition* $y(a)=b$. Depending on whether or not a>-2, and on whether or not b >-1, you may need to make some sign changes in your solution, since for $w$ negative, there is no real number $r$ such that $r=\ln w$, but $\int \frac{dw}{w}$ makes sense for w<0.2011-12-27
3

As I said in my answer to your last question, what we need when solving a differential equation (a simple example of which is taking a definite integral) is a constant of integration that can take on any real value. For this reason, we cannot write $A^2$ as our constant of integration, because we can never get a negative value for $A^2$ by plugging in a real value for $A$. However, we can write $\ln(A)$ for our constant because for any real number $r$ we get our constant of integration equal to $r$ when we plug in $e^r$ for $A$.

As for why using $\ln(A)$ as the constant of integration is more convenient than using $C$, its just so that you can write $(1+y)=A(2+x)$ rather than $(1+y)=e^C(2+x)$, and whoever wrote your textbook decided the first was a nicer expression.