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By taking 1st and 2nd equations 1st solution is $$ x=y+c_{0}$$ but i am getting two different solutions slightly differing :

Solution 1 : using lagrange multiplier 1,1,0 we get $$ \frac{dx+dy}{2z}=\frac{(3z+1)dz}{x+y}$$ $$ \frac {(x+y)d(x+y)}{2z}= {(3z+1)(dz)}$$ $$ \frac {d(x+y)^{2}}{2} = (2z)(3z+1)(dz) =(6z^{2}+2z)dz$$ $$ \frac {(x+y)^2}{2} = {2z^3}+(z^2)+c_{1} $$ $$ (x+y)^{2}= 4z^{3}+2z^{2}+c_{1'} \to(1)$$

Solution 2 : taking x,y,-z as multipliers

$$ \frac {xdx+ydy-(zdz)(3z+1)}{xz+yz-z(x+y)} =\frac {xdx+ydy-(zdz)(3z+1)}{0}$$

$$ xdx+ydy= (3z+1)(zdz)=(3z^{2}+z)dz$$ $$ \frac{x^2}{2} +\frac{y^2}{2}= z^{3}+ \frac{z^2}{2}+c_{2} $$ $$ x^{2}+y^{2}=2z^{3}+z^{2}+c_{2'}$$ $$ 2(x^{2}+y^{2})=4z^{3}+2z^{2}+c_{2''} \to(2)$$

here the first solution $(1)$ is different and the term of $$xy$$ is extra .. i believe i have done mistake in my first solution but i am unable to find out where?

solution $(1)$ is my solution and solution $(2)$ is the solution from the the institute

lol
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1 Answers1

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There is no mistake in your calculus.

Both : $$ (x+y)^{2}= 4z^{3}+2z^{2}+c_{1'} \to(1)$$ $$ 2(x^{2}+y^{2})=4z^{3}+2z^{2}+c_{2''} \to(2)$$ are correct, each one on different characteristic curves (because different constants of integrations).

This is not contradictory (on the characteristic curves, but not elsewhere) because $$4z^3+2z^2=(x+y)^2-c_{1''}=2(x^2+y^2)-c_{2''}$$ $$2(x^2+y^2)-(x+y)^2=c_{2''}-c_{1''}$$ $$x^2+y^2-2xy=c_{2''}-c_{1''}$$ $$(x-y)^2=c_{2''}-c_{1''}$$ Since $x-y=c_0$ $$c_0^2=c_{2''}-c_{1''}$$ This denotes the relationship between the constants of integrations as expected.

JJacquelin
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