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To Solve: $$\displaystyle x(z-2y^2)\frac{\partial z}{\partial x}=\left(z-\frac{\partial z}{\partial y}\right)(z-y^2-2x^3)$$

Forming the subsidiary equations: $\displaystyle \frac{dx}{x(z-2y^2)}=\frac{dy}{yz-y^3-2x^3y}=\frac{dz}{z^2-y^2z-2x^3z}$

My Attempt: $\displaystyle \frac{dy}{y(z-y^2-2x^3)}=\frac{dz}{z(z-y^2-2x^3)}$

$\displaystyle \frac{dy}{y}=\frac{dz}{z}$

$\displaystyle \frac{y}{z}=C_1 ; d\left(\frac{y}{z}\right)=0$

I can't think of the second part .. Please help.

The given answer is: $\displaystyle f\left(\frac{y}{x},\frac{z}{x}-\frac{y}{x}+x^2\right)=0$

Shaun
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square_one
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1 Answers1

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Please take a look at the solution of this question and this question

The goal is to prove that $$d\left(\frac{y}{x}\right)=0,d\left(\frac{z}{x}-\frac{y}{x}+x^2\right)=0$$

Community
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mike
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  • I got $\displaystyle d\left(\frac{y}{z}\right)=0$ .. but answer requires $\displaystyle d\left(\frac{y}{x}\right)=0$ .. What's wrong here ? – square_one Jun 21 '14 at 09:29