if y=4 when x=1, what's the particular solution to the differential equation (dy/dx)=(1/x)+(1/x^2)?

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if y=4 when x=1, what's the particular solution to the differential equation (dy/dx)=(1/x)+(1/x^2)?

2007-05-05 08:27:07 · 2 answers · asked by fawrkgn;ljn k 1 in Science & Mathematics ➔ Mathematics

2 answers

From your previous question at http://answers.yahoo.com/question/index?qid=20070505122608AASZqZl we have:
y = ln(x) - 1/x + C

Thus from
4 = ln(1) - 1 + C
C = 5
Thus,
y = ln(x) - 1/x + 5

2007-05-05 08:31:16 · answer #1 · answered by Deriver 3 · 1⤊ 0⤋

dy=1/x dx+(1/x^2)dx

integrate

y=lnx-1/x+C

4=ln1-1+C

C=5

y=lnx-(1/x)+5

2007-05-05 15:37:40 · answer #2 · answered by iyiogrenci 6 · 0⤊ 0⤋