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$$\left(1-x^2\right) y''-4 x y'-\left(1+x^2\right) y=x $$

I am required to solve the above differential equation. Can't get around how to approach. Any help would be appreciated. $y' = \frac{dy}{dx}$

Claude Leibovici
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T.Pal
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  • Welcome to MSE community! We do not encourage plaintext math formulas on this site, it will be much better if you can format your math contents in MathJax. See here for a quick guide. – Vim Mar 22 '16 at 09:02
  • I edited your equation. Check that it is exactly the one you need to solve. – Claude Leibovici Mar 22 '16 at 09:18

2 Answers2

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Hint

Try $$y(x)=\frac {z(x)}{1-x^2}$$ This will lead to a very simple differential equation in $z(x)$.

Claude Leibovici
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  • Thank you. But I can't see why this is simple. Just differentiating twice this function one get a kilometric expression, even after simplifying, and substituting in the original equation doesn't seem to render something very nice. – DonAntonio Mar 22 '16 at 09:21
  • @Joanpemo. It is not kilometric and it simplifies to something very very simple. – Claude Leibovici Mar 22 '16 at 09:24
  • Thank you very much. I will wait, perhaps the asker already got it. I tried and it became even longer than the original one. – DonAntonio Mar 22 '16 at 09:33
  • Because of your comment, I did it again and I confirm. Cheers. – Claude Leibovici Mar 22 '16 at 09:37
  • I couldn't think of this substitute. Works well. – T.Pal Mar 22 '16 at 10:25
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    In fact, knowing the result of the first substitution, the best substitution would have been $y(x)=\frac {z(x)+x}{1-x^2}$ (hard to find from scratch) leading to $z''(x)+z(x)=0$. – Claude Leibovici Mar 23 '16 at 04:36
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The method is the same as Claude Leibovici suggested, but maybe you should rather try the the following

$$z(x)=(1-x^2)y(x)$$ $$z'(x)=-2xy(x)+(1-x^2)y'(x)$$ $$z''(x)=-2y(x)-2xy'(x)-2xy'(x)+(1-x^2)y''(x)$$ $$z''(x)=(1-x^2)y''(x)-4xy'(x)-2y(x)$$

Adding $$y(x)(1-x^2)$$ on both sides results in

$$z''(x)+y(x)(1-x^2)=(1-x^2)y''(x)-4xy'(x)-(1+x^2)y(x)$$

Note that $y(x)(1-x^2)=z(x)$ and the right hand side is equal to $x$. Hence, you get the simple equation.

$$z''(x)+z(x)=x$$

Can you continue from here?

MrYouMath
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  • This simplifies. Great. – T.Pal Mar 22 '16 at 10:24
  • Elegant approach ! – Claude Leibovici Mar 22 '16 at 11:48
  • @ClaudeLeibovici: Sorry, i didn't want to copy you i just wanted to show that your suggestion really leads to an easier expression, without differentiating $z(x)/(1-x^2)$. If you feel like i copied your work, i can delete my answer. – MrYouMath Mar 22 '16 at 12:35
  • I really meant that your solution is very elegant and this is the truth ! I never thought that you copied me on any manner. You please do not delete your answer. Moverover, I upvoted your answer as soon as it came. Again, good answer, my friend ! – Claude Leibovici Mar 22 '16 at 17:10