Second order differential equation

**gundu24** · February 13th 2010, 09:03 PM

Hi everybody, I have absolutely no idea how to tackle this problem:

Let p(x) be a solution of the differential equation $y\prime\prime + yy\prime = x^{3}$ with initial conditions $y(-1) = 1$ and $y\prime(-1)=2$ . Find $p\prime\prime(-1)$ and $p\prime\prime\prime(-1)$ .

Any help would be greatly appreciated.
Thank you!

**HallsofIvy** · February 14th 2010, 03:45 AM

Originally Posted by gundu24

Hi everybody, I have absolutely no idea how to tackle this problem:

Let p(x) be a solution of the differential equation $y\prime\prime + yy\prime = x^{3}$ with initial conditions $y(-1) = 1$ and $y\prime(-1)=2$ . Find $p\prime\prime(-1)$ and $p\prime\prime\prime(-1)$ .

Any help would be greatly appreciated.
Thank you!

Did you try anything on this?

Since p(x) satisfies the equation, $p''(x)= x^3- p(x)p'(x)$ . You know the values of x, y, and y' at -1 so you can directly calculate the right side.

To find p'"(-1), Differentiate both sides of the equation above and evaluate at x=-1.

**chisigma** · February 15th 2010, 01:44 AM

The problem is quite interesting because, applying the procedure described by HalsofIvy, You can iteratively compute the quantities $p^{(n)} (-1)$ and that allows You to obtain the Taylor expansion of $p(x)$ around $x=-1$ ...

$p(x)= \sum_{n=0}^{\infty} \frac{p^{(n)} (-1)}{n!}\cdot (x+1)^{n}$ (1)

Kind regards

$\chi$ $\sigma$

**HallsofIvy** · February 15th 2010, 02:53 AM

Yes, that's a standard, though very slow and not very practical, method of finding power series solutions to differential equations.

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