Solution. We write the equation as
(x2+y2)dy=xydx
−xydx+(x2+y2)dy=0 Consider the equation as
M(x,y)dx+N(x,y)dy=0. Find
δyδM=−x
δxδN=2xδyδM=δxδN Find the value
−M(x,y)δyδM−δxδN=−−xy−x−2x=−y3. The resulting function depends only on y. Therefore, find the integrating factor using the equation
k(y)k′(y)=−y3. Hence
k(y)=y31 .
Get the total (exact) differential equation
−y2xdx+(y3x2+y1)dy=0.Let a solution to the equation be U(x,y). Hence
δxδU=−y2x Get
U=−2y2x2+C(y) where C(y) is function of y. Therefore
δyδU=y3x2+C′(y)=y3x2+y1
C′(y)=y1С(y)=ln(y)+C where C is constant. Hence
U(x,y)=−2y2x2+ln(y)+C=k,
where k is constant.
Answer:
U(x,y)=−2y2x2+ln(y)=D where D is constant.
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