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Math Help - Solving y' -x^2 -e^y = 0 using series

  1. #1
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    Solving y' -x^2 -e^y = 0 using series

    Hello,

    I need to solve the differential equation y'-x^2-e^y=0 using series, with initial conditions x=0 and y=0.

    My problem is e^y. I know what I could have done with e^x (replace e^x, y and y' par their series in the differential equation) ... but what can I do with e^y ??

    Thanks to everybody,
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  2. #2
    MHF Contributor chisigma's Avatar
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    Lets expand y in MacLaurin series...

     y(x) = y(0) + y^{'} (0)\frac{x}{1!} + y^{''} (0)\frac{x^{2}}{2!} + y^{'''} (0) \frac{x^{3}}{3!} + \dots (1)

    The 'initial condition' supplies...

    y(0) = 0 (2)

    The DE supplies y^{'} (0) because is...

    y^{'} = e^{y} + x^{2} \rightarrow y^{'} (0) = 1 (3)

    Now we compute from (3)...

    y^{''} = e^{y} y^{'} + 2 x \rightarrow y^{''} (0) = 1 (4)

    ... and from (4)...

    y^{'''} = e^{y} y^{' 2} + e^{y} y^{''} + 2 \rightarrow y^{'''} (0) = 4 (5)

    If necessary we can perform further steps. The series expansion of y is obtained from (1) and is...

    y(x) = x + \frac{1}{2} x^{2} + \frac{2}{3} x^{3} + \dots (6)

    Kind regards

    \chi \sigma
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  3. #3
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    Thanks a lot Chi Sigma!

    I failed to see a pattern emerging so I computed the next two terms:

    <br />
y^{(4)} = <br />
3e^{y}y^{'}y^{''} <br />
+ e^{y} y^{' 3} <br />
+ e^{y} y^{'''}<br />
\rightarrow y^{(4)} (0) = 8<br />

    and

    <br />
y^{(5)} = <br />
e^{y}y^{' 4}<br />
+6e^{y}y^{' 2}y^{''}<br />
+3e^{y}y^{'' 2}<br />
+4e^{y}y^{'}y^{'''}<br />
+e^{y}y^{(4)}<br />
\rightarrow y^{(5)} (0) = 34<br />

    (I used Maple for the latest one)

    Therefore we have:
    <br />
y(x) = \frac{x}{1!} + \frac{x^{2}}{2!} <br />
+ 4 \frac{x^{3}}{3!} <br />
+ 8 \frac{x^{4}}{4!} <br />
+ 34 \frac{x^{5}}{5!} <br />
+ \dots<br />

    or

    <br />
y(x) = x + \frac{1}{2} x^{2} <br />
+ \frac{2}{3} x^{3} <br />
+ \frac{1}{3} x^{4} <br />
+ \frac{17}{60} x^{5} <br />
+ \dots<br />

    Is there any pattern in that? Because if I don't have the general term how can I say I solved the DE ?

    Thanks for your help,
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