Conceptual Second order differential eqn question

by CAF123
Tags: conceptual, differential, order
 P: 1,807 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?
Math
Emeritus
Thanks
PF Gold
P: 38,460
 Quote by 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.
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.

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

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

 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?
Yes, "particular integral" is a better term than "particular solution".
P: 1,807
 Quote by HallsofIvy 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.
Should that be the 'associated homogeneous' eqn?

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

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

P: 1,807

Conceptual Second order differential eqn question

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