Question #88345
Solve the following differential equations:
i) (x²+y²)dy/dx=xy
1
Expert's answer
2019-04-28T17:35:02-0400

Solution. We write the equation as


(x2+y2)dy=xydx(x^2+y^2)dy=xydx

xydx+(x2+y2)dy=0-xydx+(x^2+y^2)dy=0

Consider the equation as


M(x,y)dx+N(x,y)dy=0.M(x,y)dx+N(x,y)dy=0.

Find


δMδy=x\frac {\delta M} {\delta y} = - x

δNδx=2x\frac {\delta N} {\delta x} = 2xδMδy≠δNδx\frac {\delta M} {\delta y} =\not \frac {\delta N} {\delta x}

Find the value


δMδyδNδxM(x,y)=x2xxy=3y.-\frac {\frac {\delta M} {\delta y}-\frac {\delta N} {\delta x}} {M(x,y)} =-\frac {-x-2x} {-xy} =- \frac {3} {y}.

The resulting function depends only on y. Therefore, find the integrating factor using the equation


k(y)k(y)=3y.\frac {k'(y)} {k(y)} = -\frac {3} {y}.

Hence


k(y)=1y3k(y)=\frac {1} {y^3}

.

Get the total (exact) differential equation


xy2dx+(x2y3+1y)dy=0.-\frac {x} {y^2}dx+(\frac {x^2} {y^3}+\frac {1} {y})dy=0.

Let a solution to the equation be U(x,y). Hence


δUδx=xy2\frac {\delta U} {\delta x} = -\frac {x} {y^2}

Get


U=x22y2+C(y)U=- \frac {x^2} {2y^2} +C(y)

where C(y) is function of y. Therefore


δUδy=x2y3+C(y)=x2y3+1y\frac {\delta U} {\delta y} =\frac {x^2} {y^3}+C'(y)=\frac {x^2} {y^3}+\frac {1} {y}

C(y)=1yC'(y)=\frac {1} {y}С(y)=ln(y)+CС(y)=ln(y) + C

where C is constant. Hence


U(x,y)=x22y2+ln(y)+C=k,U(x,y)=- \frac {x^2} {2y^2}+ln(y)+C=k,

where k is constant.

Answer: 

U(x,y)=x22y2+ln(y)=DU(x,y)=- \frac {x^2} {2y^2}+ln(y)=D

where D is constant.


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