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Conceptual Second order differential eqn question 
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#1
Nov2212, 09:49 AM

PF Gold
P: 2,305

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? 


#2
Nov2212, 11:27 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,495




#3
Nov2212, 12:09 PM

PF Gold
P: 2,305




#4
Nov2412, 11:55 AM

PF Gold
P: 2,305

Conceptual Second order differential eqn question
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