Question #296627

determine the general/particular solution for each equation using the applicable solution to equations of order one (separable, homogenous,linear,exact)


3x(xy-2)dx + (x^3=2y)dy = 0


1
Expert's answer
2022-02-14T10:48:00-0500

Let us determine the general solution of the differential equation


3x(xy2)dx+(x32y)dy=0.3x(xy-2)dx + (x^3-2y)dy = 0.


Taking into account that


(3x(xy2))y=3x2=(x32y)x,\frac{\partial( 3x(xy-2))}{\partial y} =3x^2=\frac{\partial (x^3-2y)}{\partial x},


we conclude that this equation is exact, and hence there is a function u=u(x,y)u=u(x,y) such that


ux=3x(xy2)=3x2y6x,  uy=x32y.\frac{\partial u}{\partial x}=3x(xy-2)=3x^2y-6x,\ \ \frac{\partial u}{\partial y}=x^3-2y.


It follows that u=x3y3x2+c(y),u=x^3y-3x^2+c(y), and thus


uy=x3+c(y)=x32y.\frac{\partial u}{\partial y}=x^3+c'(y)=x^3-2y.


Therefore, c(y)=2y,c'(y)=-2y, and thus c(y)=y2+C.c(y)=-y^2+C.


We conclude that the general solution is of the differential equation


x3y3x2y2=C.x^3y-3x^2-y^2=C.


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