I am having some difficulty having a more deep understanding of solving general ODE,

I will give an example because it is hard to explain without me showing some work.

Find the general solution to the ODE;


So, what I did was first considered the characteristic equation of the associated homogenous ODE,

ie $$r^{2}-2r-3=0$$


giving $r_1=-1$ and $r_2=3$ which is two real distinct roots, therefor we will have a complementary solution of the form $$y_c=c_1e^{3t}+c_2e^{-t}$$ with $c_1,c_2 \in \mathbb{R}$

Now here is where my issue begins, so I want to use undetermined coefficients

I was taught that here it is advised to look at $g(x)=3e^{2t}$ and consider what the form of the particular solution may be, that is $$y_p(t)=At^{s}e^{kt}$$ where s is the smallest non zero integer that assures we don't have our particular solution to be a linear combination of our complementary solution.

However, when I first see this, it would seem to me that I would need to take s=1 as by looking at the complementary solution it looks like it would be a linear combination. However, I checked the wronskian to see that actually $\{e^{3t},e^{-t},e^{2t}\}$ is linearly independent. So is that why I would take s here to be zero.

So in order to find s for these type of problems, including those of the form $t^{s}(Asin(t)+Bcos(t))$ etc, is it required that I start with s=0? and test the Wronskian with the fundamental solutions, and increase s by 1 until we find that specific s which is largest and gives us a 0 wronskian?

Thank you a lot, I really hope I can get some help here.

  • $\begingroup$ to find a particular solution of $y,$ it would be whole lot easier to make a change of variable $u = ye^{-2t}, y = ue^{2t}.$ then find and solve an easier equation for $u.$ $\endgroup$ – abel May 15 '15 at 2:31

:As the right-hand side has a standard form, it is much simpler: as $2$ is not a root of the characteristic equation, a particular solution has the form $A\mathrm e^{2t}$.

More genertally, if the r.h.s. is $p(t)\mathrm e^{2t}$, where $p$ is a polynomial of degree $d$, a particular solution would be $q(t)\mathrm e^{2t}$, where $q(t)$ is polynomùial of degree $d$.

If $2$ were a simple root of the characteristic equation, a particular solution would have the form $tq(t)\mathrm e^{2t}\enspace(\deg q=d)$.

If it were a double root of the characteristic equation, a particular solution would be $t^2q(t)\mathrm e^{2t}$.

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  • $\begingroup$ Thanks, what is the definition of simple root? $\endgroup$ – Quality May 15 '15 at 5:34
  • $\begingroup$ A general answer (not only for quadratic equation, a polynomial $p(x)$ has $2$ as a double root if it can be factored as $(x-2)^2q(x)$, where $2$ is not a root of $q(x)$. This is equivalent to $p(2)=p'(2)=0$ and $p''(2)\ne0 $. For a quadratic equation equation, it is the same as $p(2)=0$ and $\Delta=0$. $2$ is a simple root of $p(x)$ if $p(x)$ can be factored as $(x-2)q(x)$, where $2$ is not a root of $q(x)$. This is equivalent to $p(2)=0,\ p'(2)\ne0 $. For a quadratic equation, it means $p(2)=0$ and $\Delta\ne0$. $\endgroup$ – Bernard May 15 '15 at 7:49
  • $\begingroup$ Thanks that helped, if you have time can you please tell me if this is also correct then; if we were to consider $y''-2y'-3y=-3te^{-t}$ for example. since the char equation has roots -1 and 3 , we would have to cover up to t^2? Because the -1 appears on the e, but does the sign of the 3 matter? $\endgroup$ – Quality May 17 '15 at 21:00
  • $\begingroup$ Since $-1$ is a simple root of the characteristic equation, you would try $t(at+b)\mathrm e^{-t}$ as a solution. $\endgroup$ – Bernard May 17 '15 at 21:07
  • $\begingroup$ Thanks , and Im just making sure I understand lets say you had some equation$ y''-ay'+by=t^{2}e^{-3t}$ and just for sake of the example say that the roots of the characteristic equation were both -3, then you would try $t^2(At+b)e^{-3t}$ or something ? $\endgroup$ – Quality May 17 '15 at 21:18

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