# Second order differential equation

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• Feb 13th 2010, 09:03 PM
gundu24
Second order differential equation
Hi everybody, I have absolutely no idea how to tackle this problem:

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

Any help would be greatly appreciated.
Thank you!
• Feb 14th 2010, 03:45 AM
HallsofIvy
Quote:

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 $\displaystyle y\prime\prime + yy\prime = x^{3}$ with initial conditions $\displaystyle y(-1) = 1$ and $\displaystyle y\prime(-1)=2$. Find $\displaystyle p\prime\prime(-1)$ and $\displaystyle p\prime\prime\prime(-1)$.

Any help would be greatly appreciated.
Thank you!

Did you try anything on this?

Since p(x) satisfies the equation, $\displaystyle p''(x)= x^3- p(x)p'(x)$. 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.
• Feb 15th 2010, 01:44 AM
chisigma
The problem is quite interesting because, applying the procedure described by HalsofIvy, You can iteratively compute the quantities $\displaystyle p^{(n)} (-1)$ and that allows You to obtain the Taylor expansion of $\displaystyle p(x)$ around $\displaystyle x=-1$ ...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Feb 15th 2010, 02:53 AM
HallsofIvy
Yes, that's a standard, though very slow and not very practical, method of finding power series solutions to differential equations.