#Second order differential equation

16 messages · Page 1 of 1 (latest)

bitter veldt
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So if I am evaluating such an equation using the method of variable parameters, if I switch y1 and y2
(Basically from the auxiliary equation, using roots, y1 and y2 to form the CF)
Why does it change the particular integral?
And how can I tell which answer is valid or both are valid

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mild terrace
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Well if for example x1 and x2 are the two solutions that generate the plane of homogeneous solutions using the variable of constant if you have a solution of the form Ax1+Bx2 with A and B both class C1 functions you should get A’x1+B’x2=0 and A’x1’+B’x2’=r with r the secondary member of the equation here finding A and B will give you a particular solution

bitter veldt
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For example

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Now when I consider
C.F to be
C1 x^-1 + C2 x^-2
Calculate the wronskian and find u and v (using u = integration (y2X/W)) to find the particular integral
Yp = uy1 + vy2
I'm getting the correct answer
But if I swap y1 and y2 it's not giving the same answer, why is that? And is this answer incorrect

mild terrace
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Well if you have a particular solution of the form uy1+vy2 swapping y1
and y2 doesn’t give a solution

bitter veldt
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And why is that? If I started with the opposite selection?

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I happen to select that one

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But if I would have selected the alternate y1 and y2 would that be incorrect?

mild terrace
bitter veldt
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Yea sorry my bad I was very sleepy probably and confused myself due to an error in the calculation

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It's fixed now ty

mild terrace
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Okok np

bitter veldt
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+close