Submitted by Anonymous (not verified) on
Forums:
d2ydx2+n2x=0
In the FP3 textbook (AQA) is says the auxiliary equation for this is
k2+n2=0
I'm just wondering how you get to this :)
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Submitted by Anonymous (not verified) on
d2ydx2+n2x=0
In the FP3 textbook (AQA) is says the auxiliary equation for this is
k2+n2=0
I'm just wondering how you get to this :)
I differentiated with respect
Submitted by O________O (not verified) on
I differentiated with respect to x which leads to k3+n2=0
so still a bit lost.
Auxiliary equations
Submitted by Rayman (not verified) on
Substitute y=ekx into the differential equation and you get the auxiliary equation.
You can also do this for the second order differential equations with constant coefficients to get ak2+bk+c=0
Actually
Submitted by Rayman (not verified) on
y=Aekx
x and y
Submitted by O________O (not verified) on
would that not only work if the equation was this:
d2ydx2+n2y=0
My bad.
Submitted by Rayman (not verified) on
I misread the question.
For the other differential equation,
I'm fairly certain you can integrate it twice.
Could it be a typo in the textbook? Should that x have been a y?
I thought that too
Submitted by O________O (not verified) on
I thought it was a typo but I found it cropping up somewhere else as well - I doubt it will come up at all but it's a bit weird - Thanks :D
Auxiliary equation
Submitted by stcs on
Certainly a typo. The idea of an auxiliary equation only applies to linear equations with constant coefficients (or related equations). As someone above says, if the equation is really
d2ydx2=−n2x
you would just integrate twice to get y=−n2x3/6+Ax+B: no auxiliary equation required.