Exercises 2.5equation has an integrating factor depending onyalone. Also, since∂M/∂y−∂N/∂xN=(6y2+ 4y)−(3y2+ 2y)3y2x+ 2xy=3y2+ 2yx(3y2+ 2y)=1x,the equation has an integrating factor depending onx.Writing the equation in the formdxdy=−3y2x+ 2xy2y3+y2=−xy(3y+ 2)2y2(y+ 1)=−y(3y+ 2)2y2(y+ 1)xwe conclude that it is separable and linear withxas the dependent variable.3.This equation is not separable because of the factor (y2+ 2xy). 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)2y+ 2x=−2x=Nx(x, y).Therefore, the equation is not exact. To see if we can find an integrating factor of the formµ(x), we compute∂M∂y−∂N∂xN=2y+ 4x−x2,which is not a function ofxalone. To see if we can find an integrating factor of the form
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