Find the integrating factor of given differential equation
(x2y + y2) dx + (y3-x3) dy = 0
This Given Differential Equation is of the form:
M(x,y)dx+N(x,y)dy=0M(x,y)dx+N(x,y)dy=0M(x,y)dx+N(x,y)dy=0
where M(x,y)=x2y+y2M(x,y)=x^2y+y^2M(x,y)=x2y+y2 and N(x,y)=y3−x3N(x,y)=y^3−x^3N(x,y)=y3−x3 Now
Mx−Ny=(x2y+y2)x−(y3−x3)yMx−Ny=(x^2y+y^2)x-(y^3-x^3)yMx−Ny=(x2y+y2)x−(y3−x3)y
=x3y+xy2−y4+x3y=2x3y+xy2−y4≠0\\=x^3y+xy^2-y^4+x^3y\\=2x^3y+xy^2-y^4\\≠0=x3y+xy2−y4+x3y=2x3y+xy2−y4=0
Therefore the integrating factor (I.F.)=1Mx−Ny( I.F.)=\dfrac{1}{Mx-Ny}(I.F.)=Mx−Ny1
=12x3y+xy2−y4=\dfrac{1}{2x^3y+xy^2-y^4}=2x3y+xy2−y41
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