Answer:-
dxy+1=dyx+1=dzz\frac{dx}{y+1}=\frac{dy}{x+1}=\frac{dz}{z}y+1dx=x+1dy=zdz
∫(x+1)dx=∫(y+1)dy\int (x+1)dx=\int (y+1)dy∫(x+1)dx=∫(y+1)dy
x2/2+x=y2/2+y+c1x^2/2+x=y^2/2+y+c_1x2/2+x=y2/2+y+c1
dx−dyy−x=dzz\frac{dx-dy}{y-x}=\frac{dz}{z}y−xdx−dy=zdz
−ln(x−y)=ln(c2z)-ln(x-y)=ln(c_2z)−ln(x−y)=ln(c2z)
c2=1z(x−y)c_2=\frac{1}{z(x-y)}c2=z(x−y)1
F(x2/2+x−y2/2−y,1z(x−y))=0F(x^2/2+x-y^2/2-y,\frac{1}{z(x-y)})=0F(x2/2+x−y2/2−y,z(x−y)1)=0
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