How do I solve a differential equation with boundary conditions?

Ted123
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Homework Statement



[PLAIN]http://img402.imageshack.us/img402/7427/diff5.jpg

The Attempt at a Solution



How do I evaluate \frac{d}{dx} \left( \frac{y_2}{y_1} \right) ?
 
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Ted123 said:

Homework Statement



[PLAIN]http://img402.imageshack.us/img402/7427/diff5.jpg

The Attempt at a Solution



How do I evaluate \frac{d}{dx} \left( \frac{y_2}{y_1} \right) ?

Have you considered looking at the quotient rule? It seems at first glance like the numerator would be similar to W[y_1,y_2].

Then you could see that the integral of that might be a constant, and that you might be able to deduce that constant at the boundaries, and if that constant happened to be 1, then y_1 / y_2 would be 1, in which case y_1=y_2, for all x.
 
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So

y_2 = \frac{y_1 y_2^'}{y_1^'}

and

y_1 = \frac{y_2 y_1^'}{y_2^'}

So we want to find:

\frac{d}{dx} \left( \frac{y_2}{y_1} \right) = \frac{d}{dx} \left( \frac{y_1 {y_2^'}^2}{y_2 {y_1^'}^2} \right)

Letting u = y_1 {y_2^'}^2 and v = y_2 {y_1^'}^2

what is u' and v'?
 
Calculate the derivitive directly:

\frac{d}{dx}\bigg(\frac{y_2}{y_1}\bigg)=\frac{y_2&#039;y_1-y_2y_1&#039;}{y_1^2}[\latex]<br /> <br /> If we look at the numerator, it is 0. Therefore <br /> <br /> \frac{d}{dx}\bigg(\frac{y_2}{y_1}\bigg)=0[\latex]&lt;br /&gt; &lt;br /&gt; Therefore&lt;br /&gt; &lt;br /&gt; \bigg(\frac{y_2}{y_1}\bigg)=C for all x[\latex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Where C is some constant.&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Look at the boundery condition at x_0, &amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; \bigg(\frac{y_2(x_0)}{y_1(x_0)}\bigg)=1, as y_1(x_0)=y_2(x_0)[\latex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; thus C=1,&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; Thus &amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; \bigg(\frac{y_2}{y_1}\bigg)=1 for all x [\latex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; Which implies that y_1=y_2 for all x!&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; I can&amp;amp;amp;amp;#039;t get the latex to work...&amp;amp;amp;lt;br /&amp;amp;amp;gt; But I hope the point is clear.
 
Calculate the derivitive directly:

\frac{d}{dx}\bigg(\frac{y_2}{y_1}\bigg)=\frac{y_2&#039; y_1-y_2y_1&#039;}{y_1^2}

If we look at the numerator, it is 0. Therefore

\frac{d}{dx}\bigg(\frac{y_2}{y_1}\bigg)=0

Therefore

\frac{y_2}{y_1}=C=\text{const} for all x.

Where C is some constant.

Look at the boundery condition at x_0,

\bigg(\frac{y_2(x_0)}{y_1(x_0)}\bigg)=1, as y_1(x_0)=y_2(x_0)

thus C=1,

Thus

\frac{y_2}{y_1} =1 for all x

Which implies that y_1=y_2 for all x!


I can't get the latex to work...
But I hope the point is clear.

Thanks!

The latex wouldn't work as you should end it with [/itex] not [\latex] !
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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