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Thread: Linear , homogenous, separation and change of variables

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    Junior Member romeobernard's Avatar
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    Linear , homogenous, separation and change of variables

    Hi can you help me with this problem i dont know where do I start solving this one. Thanks

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    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    I would let:

    $\displaystyle u=2x-y+3\implies du=2\,dx-dy$

    Can you now transform the given ODE into a linear equation?
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    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Sir they already have the answer on the book

    y+c = -ln|2x+y+4|

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    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Well, I'm trying to help you derive the solution. And are you sure the solution given isn't:

    $\displaystyle y+C=-\ln|2x-y+4|$ ?
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    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Yes sir my bad. That the answer its negative y.

    But how to come up to a positive 4?

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    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    This is what i got sir.

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    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Quote Originally Posted by romeobernard View Post
    Yes sir my bad. That the answer its negative y.

    But how to come up to a positive 4?
    You will see why when you make the substitution and solve the resulting linear equation. Can you make the substitution I suggested and write the resulting ODE in standard linear form?
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    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Quote Originally Posted by romeobernard View Post
    This is what i got sir.
    Please, just call me Mark.

    When you found $\displaystyle du$ you made an error...check my first post to see what I wrote.
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    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Ok mark now i get it. You will get 1 from dividing dy/dy.

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    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Quote Originally Posted by romeobernard View Post
    Ok mark now i get it. You will get 1 from dividing dy/dy.
    What I wound up with is:

    $\displaystyle \frac{du}{dy}+u=-1$

    I solved it as a linear equation and got the implicit solution, after back-substitution:

    $\displaystyle 2x-y+4=c_1e^{-y}$

    This is equivalent to the solution you cited from your book.
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    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    This is what i get mark base on youre substitutions.. correct me if im wrong thanks.

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    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    This what i get.

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    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    Yes, that looks good.

    You chose to separate variables, which is fine.
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    Junior Member romeobernard's Avatar
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    Re: Linear , homogenous, separation and change of variables

    And how about this one.? Its a linear stated in the book.can help me how to figure out this one. Thanks

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    MHF Contributor MarkFL's Avatar
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    Re: Linear , homogenous, separation and change of variables

    That pic is sideways for me, but I will try my best.

    I would write in standard linear form as:

    $\displaystyle \frac{dr}{d\theta}+2\cot(\theta)r=-\csc(\theta)$

    Next, we compute the integrating factor:

    $\displaystyle \mu(\theta)=\exp\left(2\int \cot(\theta)\,d\theta\right)=\sin^2(\theta)$

    And the ODE becomes:

    $\displaystyle \sin^2(\theta)\frac{dr}{d\theta}+2\sin(\theta)\cos (\theta)r=-\sin(\theta)$

    Or:

    $\displaystyle \frac{d}{d\theta}\left(\sin^2(\theta)r\right)=-\sin(\theta)$

    Can you proceed?
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