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

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!

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What's a private solution? – Hans Lundmark Mar 24 '12 at 18:34
I meant that they denote the equation. – Jozef Mar 24 '12 at 19:47
@Hans I think he means particular solutions or something of the sort (in contrast with the general solution) – Pedro Tamaroff Mar 25 '12 at 3:21
I think it looks easier to prove that $\ln \frac{y_2-y_3}{y_3-y_1}$ is constant by derivating. – N. S. Mar 25 '12 at 4:02

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 $$\frac{A_1 e^{\int a(x)\,dx}}{A_2e^{\int a(x)\,dx}} = \frac{A_1}{A_2}.$$

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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... – N. S. Mar 25 '12 at 4:00
@N.S. Sometimes the math is ugly. – Pedro Tamaroff Mar 25 '12 at 4:11
@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? If necessary I will remove this comment. – Américo Tavares Mar 25 '12 at 22:28
@ArturoMagidin: Thanks anyway! – Américo Tavares Mar 25 '12 at 22:37

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

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Are you sure? It didn't work for me.. – Jozef Mar 24 '12 at 14:07
@Jozef: You must have made an error; the numerator cancels for me. – Arturo Magidin Mar 25 '12 at 3:36

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

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

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

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