# Conceptual Second order differential eqn question

1. Nov 22, 2012

### CAF123

I know that if $Y_1$ and $Y_2$ are two solutions of a nonhomogeneous second order differential eqn, then $Y_1 - Y_2$ is also a solution. So this motivates the following: if we set $Y_1 = y(x)$, where $y(x)$ is an arbritary soln of the nonhomogeneous ODE and $Y_2 = y_p(x)$, some particular soln, we get that $Y_1 - Y_2$ is a solution to the corresponding complementary equation, $$a(Y_1 - Y_2)'' + b(Y_1 - Y_2)' + c(Y_1 - Y_2) = 0,$$ ie $y_c(x) = Y_1 - Y_2 = y(x) - y_p(x)$.
I have two questions:
1)Why set $Y_1 = y(x)$ and $Y_2 = y_p(x)$? Could we have set $Y_1 = y_p(x)$and $Y_2 = y(x)$ so that in the end we get $y(x) = y_p(x) - y_c(x),$ where to recover the usual $y(x)= y_p(x) +y_c(x),$ we introduce an arbritary negative for the constants in the $y_c(x)$ term?

2) I often read questions: Find the general soln of ... and given the initial conditions...find the particular soln. I can do these questions fine. Conceptually though and understanding what is going on, I get a little confused here because we have already defined the term 'particular soln' (as above $y_p(x)$) in order to find the general soln. So is this two different things with the same term attached to them? I recall that $y_p(x)$ is sometimes called the particular integral?

Last edited: Nov 22, 2012
2. Nov 22, 2012

### HallsofIvy

Staff Emeritus
NO! You don't know that- it is not true. If $Y_1$ and $Y_2$ are two solutions of the same nonhomogeneous differential equation, then $Y_1- Y_2$ is a solution to the associated non-homogeneous equation.

Okay, now that is true- and is not what you said above.

I don't see any difference. You are just changing which function you call $Y_1$ and which you call $Y_2$.

Yes, "particular integral" is a better term than "particular solution".

3. Nov 22, 2012

### CAF123

Should that be the 'associated homogeneous' eqn?

Changing the order will mean that the general solution is the complementary function - the particular integral rather than the complementary function + the particular integral, no?(since it is strictly $Y_1 - Y_2$)

Ok, thanks. So $y_p(x)$ should really be called particular integral so as to avoid confusion. Where did the 'integral' come from in its name?

4. Nov 24, 2012

### CAF123

Can you help?