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Math Help - Initial value problem

  1. #1
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    Question Initial value problem

    Hello all,

    Can you see if this makes sense please.

    \frac{dy}{dx} = \frac{e^{-4x} - x}{e^{-4x} + 2x^{2} + 1} when y(0)=4

    \int \frac{f'(x)}{f(x)}=ln(f(x))+c

    -\frac{1}{4} \int \frac{4e^{-4x} + 4x}{e^{-4x} + 2x^{2} +1} dx

    y=\frac{1}{4} ln (e^{-4x} + 2x^{2} +1) + c

    y(0)=4

    4=(1)C=4

    y=\frac{1}{4} ln (e^{-4x} + 2x^{2} + 1) + 4

    I'm not sure but I think it all goes wrong when I put all equal to y.
    Would appreciate any help please.

    Thanks in advance.
    Last edited by bigroo; December 7th 2009 at 08:11 AM.
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  2. #2
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    Quote Originally Posted by bigroo View Post
    Hello all,

    Can you see if this makes sense please.

    \frac{dy}{dx} = \frac{e^{-4x} - x}{e^{-4x} + 2x^{2} + 1} when y(0)=4

    \int \frac{f'(x)}{f(x)}=ln(f(x))+c

    -\frac{1}{4} \int \frac{4e^{-4x} + 4x}{e^{-4x} + 2x^{2} +1} dx

    y=\frac{1}{4} ln (e^{-4x} + 2x^{2} +1) + c

    y(0)=4

    4=(1)C=4

    y=\frac{1}{4} ln (e^{-4x} + 2x^{2} + 1) + 4

    I'm not sure but I think it all goes wrong when I put all equal to y.
    Would appreciate any help please.

    Thanks in advance.
    It looks like some sign problems here. Differentiate y and see if you get back your DE.
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  3. #3
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    Thanks Danny,

    I'm sure it's a sign issue too but I'm pulling my hair out. Where do you think the error occurs?

    I can't see anything online to help with initial value problems so any help or solutions & their stages will be gratefully appreciated.
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  4. #4
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    Quote Originally Posted by bigroo View Post
    Thanks Danny,

    I'm sure it's a sign issue too but I'm pulling my hair out. Where do you think the error occurs?

    I can't see anything online to help with initial value problems so any help or solutions & their stages will be gratefully appreciated.
    If you put in an extra step it might help

    Quote Originally Posted by bigroo View Post
    Hello all,

    Can you see if this makes sense please.

    \frac{dy}{dx} = \frac{e^{-4x} - x}{e^{-4x} + 2x^{2} + 1} when y(0)=4

    \int \frac{f'(x)}{f(x)}=ln(f(x))+c

    -\frac{1}{4} \int \frac{4e^{-4x} + 4x}{e^{-4x} + 2x^{2} +1} dx

    y=\frac{1}{4} ln (e^{-4x} + 2x^{2} +1) + c

    y(0)=4

    4=(1)C=4

    y=\frac{1}{4} ln (e^{-4x} + 2x^{2} + 1) + 4

    I'm not sure but I think it all goes wrong when I put all equal to y.
    Would appreciate any help please.

    Thanks in advance.
    Your separated DE is y = \int \frac{e^{-4x} - x}{e^{-4x} + 2x^{2} + 1} \,dx

    If you let u = e^{-4x} + 2x^2 + 1 then

    du = -4\left(e^{-4x} - x\right)\,dx

    and so

     <br />
y = -\frac{1}{4} \int \frac{du}{u} + c<br />

    integrate, then back substitute. Then you can use your IC to find c.
    Last edited by Jester; December 7th 2009 at 11:09 AM. Reason: corrected typo
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  5. #5
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    Why would you let u = e^{-4x} minus  2x^2 + 1 when it is a +

    Is this an error or correct? Thanks
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  6. #6
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    Quote Originally Posted by bigroo View Post
    Why would you let u = e^{-4x} minus  2x^2 + 1 when it is a +

    Is this an error or correct? Thanks
    Sorry - it's a typo. It should be plus.
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  7. #7
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    Quote Originally Posted by Danny View Post
    Sorry - it's a typo. It should be plus.
    haha no worries, trying to confuse me even more!

    y=-\frac{1}{4}ln(u)+c

    y=-\frac{1}{4}ln(e^{-4x} + 2x^{2} + 1)+c

    y(0)=4

    c=\frac{1}{4}ln(2)+4

    So this makes:

    y=-\frac{1}{4}ln(e^{-4x} + 2x^{2} + 1)+\frac{1}{4}ln(2)+4

    Is this correct now?
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  8. #8
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    Quote Originally Posted by bigroo View Post
    haha no worries, trying to confuse me even more!

    y=-\frac{1}{4}ln(u)+c

    y=-\frac{1}{4}ln(e^{-4x} + 2x^{2} + 1)+c

    y(0)=4

    c=\frac{1}{4}ln(2)+4

    So this makes:

    y=-\frac{1}{4}ln(e^{-4x} + 2x^{2} + 1)+\frac{1}{4}ln(2)+4

    Is this correct now?
    Yes - it looks real good!
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  9. #9
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    Great stuff, I've added a thanks for all of your inspiration.
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