# Find a 3rd degree differential equation

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

• When you want an answer I suggest you to use mathjax here! – Fakemistake Dec 30 '18 at 19:06
• Is your $y_3(x)=e^{2x}$ the particular solution to the homogenous equation? – Arthur Dec 30 '18 at 19:16
• 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. – 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 $$y'''-2y''=\frac2{x^3}+\frac2{x^2}$$
• How did you get the $1$ in $P(D)=\{1,x,e^{2x}\}$? – Shubham Johri Dec 31 '18 at 4:35