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Math Help - Simple ODE

  1. #1
    No one in Particular VonNemo19's Avatar
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    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)
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  2. #2
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    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).
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    Quote Originally Posted by VonNemo19 View Post
    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)
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dwsmith View Post
    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...
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Jose27 View Post
    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.
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  6. #6
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    Quote Originally Posted by VonNemo19 View Post
    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
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dwsmith View Post
    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.
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  8. #8
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    Quote Originally Posted by VonNemo19 View Post
    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.
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  9. #9
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    Quote Originally Posted by VonNemo19 View Post
    So, what about going this route...

    Thanks.
    What method is that or from what page?
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dwsmith View Post
    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.
    Quote Originally Posted by dwsmith View Post
    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!
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