Solving $3y^2\frac{\partial z}{\partial x}-\left(z-x\right)\frac{\partial z}{\partial y}=3y^2$, I keep getting the same wrong answer

Question

For the conditions $x\left(t\right)=t^{\frac{1}{2}},\:y\left(t\right)=t^{\frac{1}{3}},\:z\left(t\right)=0,\:t>0$

I get this:

$$\frac{dx}{3y^2}=\frac{dy}{-z+x}=\frac{dz}{3y^2}$$

$$(I)\:\:\: \frac{dx}{3y^2}=\frac{dz}{3y^2} $$

$$x=z+C_1$$

$$C_1 = x-z$$

$$(II)\:\:\: \frac{dy}{-z+x}=\frac{dz}{3y^2} $$

$$3y^2dy\:=\:\left(-z+x\right)dz$$

$$y^3=-\frac{z^2}{2}+zx+C_2$$

$$C_2=y^3+\frac{z^2}{2}-zx$$

Putting in conditions

$$C_1=t^{\frac{1}{2}}$$

$$C_2=t$$

$$C_2=C_1^2$$

This means that

$$y^3+\frac{z^2}{2}-zx = \left(x-z\right)^2$$

$$2y^3+z^2-2zx = 2 \left(x-z\right)^2$$

$$2y^3+\left(z^2-2zx+x^2-x^2\right)\:=\:2\:\left(x-z\right)^2$$

$$2y^3-x^2+\left(x-z\right)^2\:=\:2\:\left(x-z\right)^2$$

$$2y^3-x^2=\:\left(x-z\right)^2$$

$$\sqrt{2y^3-x^2}=x-z$$

$$x-\sqrt{2y^3-x^2}=z$$

But it's still wrong

So where's the mistake?

Answer 1

$$\frac{dy}{-z+x}=\frac{dz}{3y^2}$$ You should use the constant $C_1$ instead. Since $C_1=x-z$. Your integration of this differential equation is not correct. $$3y^2{dy}=C_1{dz}$$ Integrate. $$y^3=C_1z+C_2$$ $$y^3=(x-z)z+C_2$$ $$C_2=y^3-xz+z^2$$

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This line seems not correct to me: $$y^3=-\frac{z^2}{2}+zx+C_2$$

Answer 2

$$3y^2\frac{\partial z}{\partial x}-\left(z-x\right)\frac{\partial z}{\partial y}=3y^2$$ You correctly wrote the Charpit-Lagrange characteristic ODEs : $$\frac{dx}{3y^2}=\frac{dy}{-z+x}=\frac{dz}{3y^2}$$ Don't forget that those ODEs are valid only on the characteristic curves, not elsewhere. Thus solving them leads to chacracteristic curves equations. Other equations of curves are not characteristic curves.

You correctly solved $\frac{dx}{3y^2}=\frac{dz}{3y^2}$ giving the equation of a first characteristic : $$x-z=C_1$$

In order to find a second characteristic you tried to solve $\frac{dy}{-z+x}=\frac{dz}{3y^2}\quad\implies\quad 3y^2dy=\left(-z+x\right)dz$

If you suppose $z$=constant this ODE is not valid because $z=$constant is outside the characteristic curves. Thus $\int z\,dx\neq zx$ and so : $\int (-z+x)dx\neq -zx+\frac12 x^2$ as you wrote. That was the mistake.

This would be corrct in integrating on the characteristic $x-z=C_1$ that is $z=x-C_1$ : $$3y^2dy=\left(-(x-C_1)+x\right)dz$$ $$3y^2dy=C_1dz$$ $$z-\frac{y^3}{C_1}=C_2$$ The general solution of the PDE expressed on the form of implicit equation $C_2=F(C_1)$ is : $z-\frac{y^3}{C_1}=F(C_1)$ $$\boxed{z=\frac{y^3}{x-z}+F(x-z)}$$ $F$ is an arbitrary function.

Condition : $z(t)=0\:;\:x(t)=t^{1:2}\:;\:y(t)=t^{1/3}$ $$0=\frac{t}{t^{1/2}-0}+F(t^{1/2}-0)$$ Let $X=t^{1/2}$ $$F(X)=-X$$ Now the function $F(X)$ is known. We put it into the above general solution where $X=x-z$ : $$z=\frac{y^3}{x-z}-(x-z)$$ $$\frac{y^3}{x-z}-x=0$$ Solving for $z$ gives the particular solution of the ODE which satisfies the specified condition. $$z=x-\frac{y^3}{x}$$