physics=world
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1. Find a second solution of the differential eq. by using the formula.
xy" + y' = 0 ; y1 = ln(x)
2.
y2 = y1(x) ( ∫(e-∫P(x)dx) / (y1(x)2) )dx
3.
I found the p(x):
p(x) = 1/x
and then I plug in everything into the formula:
y2 = ln(x) ∫(e-∫((1)/(x))dx) / (ln(x)2 )dx
solve:
= ln(x) ( ∫(e^(-ln(x))) / (x)(ln(x)2))
= ∫ (1)/ (x)(ln(x))
I do not know if this is correct. I'm going to need some help.
the answer is y2 = 1
xy" + y' = 0 ; y1 = ln(x)
2.
y2 = y1(x) ( ∫(e-∫P(x)dx) / (y1(x)2) )dx
3.
I found the p(x):
p(x) = 1/x
and then I plug in everything into the formula:
y2 = ln(x) ∫(e-∫((1)/(x))dx) / (ln(x)2 )dx
solve:
= ln(x) ( ∫(e^(-ln(x))) / (x)(ln(x)2))
= ∫ (1)/ (x)(ln(x))
I do not know if this is correct. I'm going to need some help.
the answer is y2 = 1
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