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I am taking a course in Differential Equations and we were shown how to use the auxiliary equation to easily get the general solution for a differential equations with constant coefficients. For example: $$ y'' - 4y' + 16y = 0 $$ has the auxiliary equation: $$ m^2 - 4 + 16= 0 $$

However, I wanted to know how one can write the auxiliary equation for a DE with undetermined coefficients. For example: $$ 2x^2y'' + 5xy' + y = 0 $$

Thank you!

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    $\begingroup$ Wouldn't that be nice? $\endgroup$ – Git Gud Nov 27 '16 at 20:30
  • $\begingroup$ It is $m^2-4m+16=0$ which has two comolex roots. for the second example, it is non linear and there is no auxilary equation . $\endgroup$ – hamam_Abdallah Nov 27 '16 at 20:38
  • $\begingroup$ thank you for the answers gentlemen... I thought this was a linear DE... may I ask what makes it non linear @AbdallahHammam? $\endgroup$ – Muaz Nov 28 '16 at 4:10
  • $\begingroup$ I am going to try and solve it how you mentioned @Moo $\endgroup$ – Muaz Nov 28 '16 at 4:10
  • $\begingroup$ @Muaz : the equation if not $m^2-4+16=0$ , but is $m^2-4m+16=0$. The roots are complex : $m=2\pm i2\sqrt{3}$ so that the solutions are $y=c_1e^{2x}\sin(2\sqrt{3}x)+c_2e^{2x}\cos(2\sqrt{3}x)$. $\endgroup$ – JJacquelin Nov 28 '16 at 11:17
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To clarify the question about linearity :
Both equations

$y'' - 4y' + 16y = 0$

$2x^2y'' + 5xy' + y = 0$

are linear.

For ODEs, linear means linear relatively to $y$ , $y'$, $y''$ even if the coefficients of them are not linear.

To solve $\quad y'' - 4y' + 16y = 0\quad$ change of function : $$y(x)=c\:e^{m\:x}$$ in order to obtain $$m^2-4m+16=0$$

To solve $\quad 2x^2y'' + 5xy' + y = 0\quad$ change of function : $$y(x)=c\:x^m $$ in order to obtain $$2m(m-1)+5m+1=0$$

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  • $\begingroup$ Thank you so much! But could you please explain how the form $ y(x) = cx^m $ is understood? How does the $ x^2 $ translate to $ m(m-1) $ ? Thanks again! $\endgroup$ – Muaz Nov 30 '16 at 3:06
  • $\begingroup$ Just put $y=cx^m$ into $2x^2y''+5xy'+y=0$ and simplify. You get $2m(m-1)+5m+1=0$ any values of $c$. $\endgroup$ – JJacquelin Nov 30 '16 at 6:54
  • $\begingroup$ That is perfect. Thank you! $\endgroup$ – Muaz Dec 5 '16 at 0:35

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