calculus
posted by waqas
integrate the following
(3y7x3)dx+(7y3x7)dy=0

Steve
rearrange things a it to get
(3y7x3)+(7y3x7)y'=0
now integrate term by term
3xy7/2x^23x + 7/2y^23xy7y=0
the 3xy's go away; multiply by 2/7 to get
(x^2+6/7)+(y^22y) = 0
(x+3/7)^2+(y1)^2 = 58/49
an hyperbola opening up and down
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