Exercises 2.5equation has an integrating factor depending onyalone. Also, since∂M/∂y−∂N/∂xN=(6y2+4y)−(3y2+2y)3y2xxy=3y2yx(3y2y)=1x,the equation has an integrating factor depending onx.Writing the equation in the formdxdy=−3y2xxy2y3+y2=−xy(3y+2)2y2(y+1)=−y(3y2y2(yxwe conclude that it is separable and linear withxas the dependent variable.3.This equation is not separable because of the factor (y2xy). It is not linear because of thefactory2. To see if it is exact, we computeMy(x, y)andNx(x, y), and see thatMy(x, y)2yx6=−2x=Nx(x, y).Therefore, the equation is not exact. To see if we can Fnd an integrating factor of the formµ(x), we compute∂M∂y−∂N∂xN=2yx−x2,which is not a function ofxalone. To see if we can Fnd an integrating factor of the form
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.