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dy/dx= (4xy-2)/[3y (square)-2x (square)] xdy-ydx= root[ x(square)- y(square)]

dy/dx= (4xy-2)/[3y (square)-2x (square)]


 


xdy-ydx= root[ x(square)- y(square)]

Grade:

1 Answers

melvin davis
6 Points
10 years ago

the second one:

divide numerator and the denominator by x^2         =     d(y/x) = ((1-(y/x)^2)^(1/2))/x

take y/x = u       =         du = ((root[1-u(square)])/x)dx

now easily integrate 

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