Knowing which particular integral to use for ordinary differential equations

Normally what's on the RHS of your equation (the "forcing term" as I'll call it) will be a trig function, an exponential function or a polynomial. Usually your particular integral will simply be something of the same form.

So if the forcing term is $\cos 4x$ then you put $y_p = A\cos 4x + B\sin 4x$. If the forcing term is $e^{3x}$ then you put $y_p = Ae^{3x}$. If the forcing term is $x^2+3$ then you put $y_p=Ax^2+Bx+C$. The key things to note here are that for trig functions you need to include both sines and cosines, because of the way their derivatives work; and for polynomials you need to match the highest power of the polynomial (you don't need to go higher).

This sometimes doesn't work when the complimentary function forms part of the forcing term. So for example if your forcing term was $e^{2x}$ and your characteristic polynomial was $\lambda^2 - 3\lambda + 2 = (\lambda - 1)(\lambda - 2)$, then your complimentary function would contain an $e^{2x}$ term. In this case, you'd notice that you can't solve for the particular integral, so you have to multiply by $x$; i.e. put $y_p=Axe^{2x}$. Similarly, say you had $\sin 2x$ in your forcing term and your characteristic polynomial was $\lambda^2 + 4$, then the same thing would happen.

The very worst-case scenario is if your forcing term is, say, $e^{2x}$ and your characteristic polynomial is $\lambda^2 - 4\lambda + 4 = (\lambda - 2)^2$. Then your complimentary function would be $(Ax+B)e^{2x}$, and so even trying $y_p=Cxe^{2x}$ wouldn't work because part of the complimentary function contains it. In this case you need to go even further and put $y_p = Cx^2e^{2x}$.

Hope this helps.

Thank you, that was reallly really helpful

Just want to confirm, using my Yp(t) = 1 - 3t would not be correct if e^t is a substitution for x in the original differential equation then? (if I have understood correctly. if not, then ) I'd have to use the quadratic version, right?

Hrm? If the RHS of your equation is $1-3t$ then the correct substitution to make for the particular integral is $At+B$. I'm not sure what you're getting at with $e^t$ being a substitution for x and so on.

Basically, I was trying to solve a cauchy-euler differential equation using the substitution x = e^t so essentially solving in terms of t and then reverting back to x for the final general solution. So consequently my complimentary function has terms which involve exponentials which is why I thought I was messing up my particular part by just putting the particular integral as a linear polynomial. I'm not sure how much of that made sense but I hope it did.

Ah right, that makes sense. In that case, make the substitution so it becomes a normal linear differential equation and then forget the x terms ever existed; just look at it as a differential equation w.r.t. t and solve it that way. Then you can substitute back right at the end. I can't help you much more though, because I'm not sure what your precise equation is.

$x^2 f''(x) + 5x f'(x) - 12y = 1 - 3 log x$

This is the equation. Is using Yp as At + B reasonable after doing the whole x = e^t substitution?

Normally what's on the RHS of your equation (the "forcing term" as I'll call it) will be a trig function, an exponential function or a polynomial. Usually your particular integral will simply be something of the same form.

So if the forcing term is $\cos 4x$ then you put $y_p = A\cos 4x + B\sin 4x$. If the forcing term is $e^{3x}$ then you put $y_p = Ae^{3x}$. If the forcing term is $x^2+3$ then you put $y_p=Ax^2+Bx+C$. The key things to note here are that for trig functions you need to include both sines and cosines, because of the way their derivatives work; and for polynomials you need to match the highest power of the polynomial (you don't need to go higher).

This sometimes doesn't work when the complimentary function forms part of the forcing term. So for example if your forcing term was $e^{2x}$ and your characteristic polynomial was $\lambda^2 - 3\lambda + 2 = (\lambda - 1)(\lambda - 2)$, then your complimentary function would contain an $e^{2x}$ term. In this case, you'd notice that you can't solve for the particular integral, so you have to multiply by $x$; i.e. put $y_p=Axe^{2x}$. Similarly, say you had $\sin 2x$ in your forcing term and your characteristic polynomial was $\lambda^2 + 4$, then the same thing would happen.

The very worst-case scenario is if your forcing term is, say, $e^{2x}$ and your characteristic polynomial is $\lambda^2 - 4\lambda + 4 = (\lambda - 2)^2$. Then your complimentary function would be $(Ax+B)e^{2x}$, and so even trying $y_p=Cxe^{2x}$ wouldn't work because part of the complimentary function contains it. In this case you need to go even further and put $y_p = Cx^2e^{2x}$.

Hope this helps.