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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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    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

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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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    @ArturoMagidin: Thanks anyway!2012-03-25
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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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    @Jozef: You must have made an error; the numerator cancels for me.2012-03-25
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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. $