# Thread: Simple ODE

1. ## Simple ODE

Hi, can someone explain to me the process of dealing with the absolute value signs in this problem:

Solve the given initial value problem and give the largest interval $I$ on which the solution is defined.

$\sin(x)\frac{dy}{dx}+y\cos(x)=0$

So, I get to this point: $\ln|y|=-\ln|sin(x)|+c_1$.

Now, how do I deal with these absolute values, and what is the justification for dropping them? WolframAlppha gives $y=c\csc(x)$

2. Well, the most obvious reason would be to verify that without these we still get a solution to the DE (besides you do want to get rid of them because they threaten the differentiability of the solution).

3. Originally Posted by VonNemo19
Hi, can someone explain to me the process of dealing with the absolute value signs in this problem:

Solve the given initial value problem and give the largest interval $I$ on which the solution is defined.

$\sin(x)\frac{dy}{dx}+y\cos(x)=0$

So, I get to this point: $\ln|y|=-\ln|sin(x)|+c_1$.

Now, how do I deal with these absolute values, and what is the justification for dropping them? WolframAlppha gives $y=c\csc(x)$
What I would do is:

$\sin(x) y'+\cos(x) y=0\Rightarrow y'+\cot(x) y=0$

$\displaystyle y\exp\left(\int\cot(x) \ dx\right)=\int 0 \ dx\Rightarrow y\sin(x)=C\Rightarrow y=C\csc(x)$

4. Originally Posted by dwsmith
What I would do is:

$\sin(x) y'+\cos(x) y=0\Rightarrow y'+\cot(x) y=0$

$\displaystyle y\exp\left(\int\cot(x) \ dx\right)=\int 0 \ dx\Rightarrow y\sin(x)=C\Rightarrow y=C\csc(x)$
I'm sorry, You're using methods that I'm not familiar with. Maybe if you showed some intermediate steps...

5. Originally Posted by Jose27
Well, the most obvious reason would be to verify that without these we still get a solution to the DE (besides you do want to get rid of them because they threaten the differentiability of the solution).
I'm looking for an analytical approach here. I don't want to have to rely on check methods to know that my solution is correct.

6. Originally Posted by VonNemo19
I'm sorry, You're using methods that I'm not familiar with. Maybe if you showed some intermediate steps...
This is the one of the first methods you learn.

$y'+r(x)y=s(x)$

$\displaystyle P(x)=\exp\left(\int r(x) \ dx\right), \ Q(x)=s(x)$

$\displaystyle \int\left[\frac{d}{dx}\left[yP(x)\right]\right]dx=\int P(x)Q(x) \ dx\Rightarrow yP(x)=\int P(x)Q(x) \ dx$

Page 53 in your book

7. Originally Posted by dwsmith
This is the one of the first methods you learn.

$y'+r(x)y=s(x)$

$\displaystyle P(x)=\exp\left(\int r(x) \ dx\right), \ Q(x)=s(x)$

$\displaystyle \int\left[\frac{d}{dx}\left[yP(x)\right]\right]dx=\int P(x)Q(x) \ dx\Rightarrow yP(x)=\int P(x)Q(x) \ dx$

Page 53 in your book
Oh, right. Sorry. Integrating factor. My bad.

So, what about going this route...

$\ln|y|=-\ln|\sin(x)|+c_1$

$\Rightarrow{y}=\pm{e}^{-\ln|\sin(x)|+c_1}$

$\Rightarrow{y}=c{e}^{\ln|\sin(x)|^{-1}}$

$\Rightarrow{y}=c{e}^{\ln|\csc(x)|}$

But, $y'+\cot(x)y=0$ is of the form $y'+P(x)y=0$ and we are only seeking a solution on an interval $I$ for which the coefficient function $P(x)$ is continuous. So, we take $I=(0,\infty)$ and therefore we are justified in removing the absolute value signs, giving

${y}=c{e}^{\ln\csc(x)}$
or
${y}=c{\csc(x)}$

What about that? Page 53. Right under equation (2).

Thanks.

8. Originally Posted by VonNemo19
Oh, right. Sorry. Integrating factor. My bad.

So, what about going this route...

$\ln|y|=-\ln|\sin(x)|+c_1$

$\Rightarrow{y}=\pm{e}^{-\ln|\sin(x)|+c_1}$

$\Rightarrow{y}=c{e}^{\ln|\sin(x)|^{-1}}$

$\Rightarrow{y}=c{e}^{\ln|\csc(x)|}$

But, $y'+\cot(x)y=0$ is of the form $y'+P(x)y=0$ and we are only seeking a solution on an interval $I$ for which the coefficient function $P(x)$ is continuous. So, we take $I=(0,\infty)$ and therefore we are justified in removing the absolute value signs, giving

${y}=c{e}^{\ln\csc(x)}$
or
${y}=c{\csc(x)}$

What about that? Page 53. Right under equation (2).

Thanks.
Cotangent is discontinuous when $x=\pi k \ \ \ k\in\mathbb{Z}$.

I would use the interval $(0,\pi)$

Check out number 17 of the problems in that section. It is remarkable similar.

9. Originally Posted by VonNemo19
So, what about going this route...

Thanks.
What method is that or from what page?

10. Originally Posted by dwsmith
What method is that or from what page?
The problem is number 19 from the chapter 2 review. On page 53 right underneath equation 2, it states what interval the solution of a linear ODE will be defined.
Originally Posted by dwsmith
Cotangent is discontinuous when .

I would use the interval
Oh yeah...But since this is an IVP, we take the interval to be $(\pi,2\pi)$. I got it!