0
$\begingroup$

I'd really love your help with the following exercise.

I need to show that if $y_1, y_2, y_3$ are particular solutions of the linear equation:

$y'+a(x)y=b(x)$, so the function $$\frac{y_2-y_3}{y_3-y_1}$$ is constant.

I got that a particular solution should be of the form: $e^{-\int_{x_0}^{x}a(s)ds}c(x)$, where $c'(x)=e^{\int_{x_0}^{x}a(s)ds}b(x)$. what else should I do? How should I solve this one?

Thanks!

  • 0
    What's a private solution?2012-03-24
  • 0
    I meant that they denote the equation.2012-03-24
  • 0
    @Hans I think he means particular solutions or something of the sort (in contrast with the general solution)2012-03-25
  • 2
    I think it looks easier to prove that $\ln \frac{y_2-y_3}{y_3-y_1}$ is constant by derivating.2012-03-25

3 Answers 3

4

If $y_1$ and $y_2$ are both particular solutions to $$y' + a(x)y = b(x)$$ then $y_1-y_2$ is a solution to the associated homogeneous differential equation $$y' + a(x)y = 0.$$

Indeed, we have $$\begin{align*} (y_1-y_2)' + a(x)(y_1-y_2) &= y_1'-y_2 + a(x)y_1 - a(x)y_2\\ &= (y_1'+a(x)y_1)-(y_2'+a(x)y_2)\\ &= b(x)-b(x)=0. \end{align*}$$

In your situation, you then have the quotient of two (nonzero) solutions to the homogeneous equation $$y' + a(x)y = 0.$$

This is separable, and a solution to this equation is of the form $Ae^{\int a(x)\,dx}$ for some constant $A$. So what you have is a quotient of the form $$\frac{A_1 e^{\int a(x)\,dx}}{A_2e^{\int a(x)\,dx}} = \frac{A_1}{A_2}.$$

  • 1
    If $y_i' + a(x)y_i = 0$ then $\frac{y'_1}{y_1}-\frac{y'_2}{y_2}=0$ thus $\ln y_1-\ln y_2$ is constant... Of course this is the same as your computation, but looks nicer...2012-03-25
  • 0
    @N.S. Sometimes the math is ugly.2012-03-25
  • 0
    @Dear Arturo Magidin, I know it is not the right place because it is rather off-topic, but would you mind having a look at my last question [123632](http://math.stackexchange.com/questions/123632/divergent-series-expansion-in-aperys-proof-of-the-irrationality-of-zeta2-a)? If necessary I will remove this comment.2012-03-25
  • 0
    @ArturoMagidin: Thanks anyway!2012-03-25
3

Differentiate the given quotient and replace $y_i'$ with $-ay_i+b$ for $i=1,2,3$ and you'll see that everything cancels.

  • 0
    Are you sure? It didn't work for me..2012-03-24
  • 0
    @Jozef: You must have made an error; the numerator cancels for me.2012-03-25
3

Just thought I'd add the quotient rule worked out:

$$\left(\frac{y_2-y_3}{y_3-y_1}\right)'=\frac{(y_2-y_3)'(y_3-y_1)-(y_2-y_3)(y_3-y_1)'}{(y_3-y_1)^2}$$

$$=\frac{\big(a(x)(\color{Blue}{y_3-y_2})\big)(\color{Green}{y_3-y_1})-(\color{Blue}{y_2-y_3})\big(a(x)(\color{Green}{y_1-y_3})\big)}{(y_3-y_1)^2}=0. $$