solve the differential equation: y'+y^2sinx=0 I used the method of separation: y'=-y^2sinx dy/dx=-y^2sinx dy/-y^2=sinxdx integral of -y^-2dy=integral of sinxdx 1/y=cosx+C y=1/(cosx+C) Did I do this right?
Last edited by mr fantastic; April 10th 2011 at 08:17 PM. Reason: Re-titled.
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Originally Posted by Taurus3 solve the differential equation: y'+y^2sinx=0 I used the method of separation: y'=-y^2sinx dy/dx=-y^2sinx dy/-y^2=sinxdx integral of -y^-2dy=integral of sinxdx 1/y=cosx+C y=1/(cosx+C) Did I do this right?
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