I'm studying differential equations and in one question on a sheet the teacher gave us for practice I'm asked to find a 3rd degree linear differential equation. I know 2 particular solutions and the solution of the associated homogeneous solution.

My question is: how do I use the particular solutions, I can write a polynome using the homogeneous and another one using the particular ones. I dont know what to do from here. I have a resolution of this question, but none of my colleagues is able to explain it, they just memorized the process for the evaluation, I'm trying to understand it. Theres a photo of the resolution.

Thank you

The resolution

  • 1
    $\begingroup$ When you want an answer I suggest you to use mathjax here! $\endgroup$ – Fakemistake Dec 30 '18 at 19:06
  • $\begingroup$ Is your $y_3(x)=e^{2x}$ the particular solution to the homogenous equation? $\endgroup$ – Arthur Dec 30 '18 at 19:16
  • $\begingroup$ You should only need one particular solution. Your solution to the homogeneous equation should have three constants in it that can be used to match the initial conditions. I can't read your photo, so cannot give details. $\endgroup$ – Ross Millikan Dec 30 '18 at 19:25

The general $3^{rd}$ order linear differential equation with constant coefficients is $$y'''+ay''+by'+cy=d;a,b,c\in\Bbb R$$

$y_1(x)=x+\ln x,y_2(x)=\ln x$ are solutions, which gives $\frac2{x^3}-\frac a{x^2}+\frac bx +b+cx+c\ln x=d=\frac2{x^3}-\frac a{x^2}+\frac bx +c\ln x$, which gives $cx+b=0\forall x\therefore b=c=0$. The associated homogeneous equation is $y'''+ay''=0$ whose solution is $e^{2x}\ \therefore a=-2$. The answer is $$y'''-2y''=\frac2{x^3}+\frac2{x^2}$$

  • $\begingroup$ thank you, I tried to use a reverse method we learned in class,in which we calculate a polynome and then find the solutions of the differential equation, and it worked. because it's a physics course we can use some algorithms the teacher gives us to skip some steps $\endgroup$ – fec Dec 30 '18 at 20:34
  • $\begingroup$ How did you get the $1$ in $P(D)=\{1,x,e^{2x}\}$? $\endgroup$ – Shubham Johri Dec 31 '18 at 4:35
  • $\begingroup$ Because I have x and I know that the polynome is just D, towhich is associated the solution 1 and x $\endgroup$ – fec Jan 2 at 17:53

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