Differential Equation Question?

0 ⤊

Solve this and show work

dy/dx = (2x^2+1)/(xe^y) , x>0

2007-09-22 13:01:40 · 3 answers · asked by Lois C 1 in Science & Mathematics ➔ Mathematics

∫(e^y)dy=∫(2x+1x)dx
e^y= x^2+lnx+c
you can leave the answer like that or solved for y by using Ln

2007-09-22 13:16:32 · answer #1 · answered by Dwasa 2 · 0⤊ 0⤋

cross multiply
dy xe^y = dx (2x^2 +1)
divide left by x since x >0
e^y dy = 2x +(1/x) dx

integrate
(1/y)e^y = x^2 + ln x + C

2007-09-22 20:09:00 · answer #2 · answered by intrepid_mesmer 3 · 0⤊ 1⤋

separating variables
e^y*dy = 2x+1/x
so e^y = x^2 +ln x +c
y=ln[x^2+lnx+c]

2007-09-22 20:10:59 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋