Integrating factor for xydx+(2x2+3y2−20)dy=0

Kathy Williams

Kathy Williams

Answered

2021-12-20

Integrating factor for xydx+(2x2+3y220)dy=0

Answer & Explanation

Karen Robbins

Karen Robbins

Expert

2021-12-21Added 49 answers

Given:

xy dx +(2x2+3y220) dy =0 
m dx +n dy =0 
my=x nx=4x 
Since:

mynx 
So

 m(y)=nxmym=4xxxy=3y 
And:

 I.f=em(y) dy =e34 
I.f=e3lny=elny3 
I.f=y3

Andrew Reyes

Andrew Reyes

Expert

2021-12-22Added 24 answers

Equation is

 M(x,y) dx +N(x,y) dy =0 with M=xy,N=2x2+3y220
it is

 My=x Nx=4x,but NxMyM=3y

if only y is a factor, then the integrating factor is

 I.F.=e3lny=y3 

and generates the precise differential. 

dF(x,y)=P dx +Q dy =0, with P(x,y)=xy4,Q(x,y)=2x2y3+3y620y3
Solution 

F(x,y)=C 

by integrating the equation.

Fx=P 

wich gives 
F(x,y)=x2y42+G(y)

Using the

 Fy=Q

 G(y)=3y520y3 
i.e. G(y)=(12)y65y4+C

Hence F(x,y)=(y4)(x22+y25)=C.

RizerMix

RizerMix

Expert

2021-12-29Added 437 answers

The equation is M(x,y)dx+N(x,y)dy=0 with M=xy,N=2x2+3y220. it is My... The solution F(x,y)=C is obtained integrating the eq.

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