Asked by waqas
integrate the following
(3y-7x-3)dx+(7y-3x-7)dy=0
(3y-7x-3)dx+(7y-3x-7)dy=0
Answers
Answered by
Steve
rearrange things a it to get
(3y-7x-3)+(7y-3x-7)y'=0
now integrate term by term
3xy-7/2x^2-3x + 7/2y^2-3xy-7y=0
the 3xy's go away; multiply by 2/7 to get
-(x^2+6/7)+(y^2-2y) = 0
-(x+3/7)^2+(y-1)^2 = 58/49
an hyperbola opening up and down
(3y-7x-3)+(7y-3x-7)y'=0
now integrate term by term
3xy-7/2x^2-3x + 7/2y^2-3xy-7y=0
the 3xy's go away; multiply by 2/7 to get
-(x^2+6/7)+(y^2-2y) = 0
-(x+3/7)^2+(y-1)^2 = 58/49
an hyperbola opening up and down
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