The correct form of the given equation is:
2x(y+z2)dx=2y(y+z2)dy=z3dz Taking the first two terms
2x(y+z2)dx=2y(y+z2)dy⟹xdx=ydy Integrating both the sides, we have:
logx=logy+logc1x=yc1c1=yx⋯(1) Taking Last two terms, we have:
2y(y+z2)dx=z3dz⟹z3dy−2y(y+z2)dz=0
Divide both sides by −z3y2
−y21dzdy+zy2=−z32 Let
y1=t⟹−y21dzdy=dzdt On comparing it with Linear differential equation, we get:
P=z2,Q=z3−2
Integrating factor, I.F.=e∫P.dz=e∫z2dz=e2logz=z2
The solution is:
t×I.F.=∫I.F.×Qdz
⇒t(z2)=∫z2×z3−2dz
⇒t×z2=−2logz+c2
⇒c2=yz2+2logz The solution of the given equation is:
ϕ(c1,c2)=0⟹ϕ(yx,yz2+2logz)=0
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