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I want to solve $$yp^2=2(z+xp+yq)$$where, $$p=z_x,q=z_y$$
My attempt:

Let $f(x,y,z,p,q)=yp^2-2(z+xp+yq)$

So that $$f_x=-2p,f_y=p^2-2q,f_z=-2,f_p=2py-2x,f_q=-2y$$
As per Charpits method:$$\displaystyle \frac{dx}{f_p}=\frac{dy}{f_q}=\frac{dz}{pf_p+qf_q}=\frac{-dp}{f_x+pf_z}=\frac{-dq}{f_y+qf_z}$$
So,putting all the values and then equating second and fourth term,I get $p=c/{y^2}$ and equating fourth and fifth term gives $pq=p^3/12 +a$, where $a$ and $c $ are constants.

Then, $$dz=pdx +qdy \space \space \space .... (A)$$
Here I got stuck and I don't know how to solve(A) for $z$.

Please help me solve this problem.Any help towards this is much appreciated.

PS:I need "complete integral" of $yp^2=2(z+xp+yq)$,i.e a solution of this form:$g(x,y,z,c_1,c_2)=0$,where $c_1$ and $c_2$ are arbitrary constants.

Shivanee Gupta
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Koro
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2 Answers2

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Hint :

You got $p=\frac{c}{y^2}$. Put this in the given equation and find the value of $q$. Then use $\,dz=p\,dx+q\,dy $ and hence solve it.

Finally get , $$\,dz=\frac{c}{y^2}\,dx+\frac{c^2-2y^3z-2cxy}{2y^4}\,dy.$$ $$\implies2(y\,dz+z\,dy)=\frac{c^2}{y^3}\,dy+2c\frac{y\,dx-x\,dy}{y^2}$$ $$\implies 2\,d(zy)=\frac{c^2}{y^3}\,dy+2c\,d(x/y)$$ $$\implies2zy=-\frac{c^2}{2y^2}+2c.\frac{x}{y}+c_1$$

Koro
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Empty
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On solving 4th and 5th term $\frac{-dp}{f_x+pf_z}$=$\frac{-dq}{f_y+qf_z}$ $\implies\frac{-dp}{-2p+p(-2)}=\frac{-dq}{p^2-2q+q(-2)}$ $\implies\frac{-dp}{-4p}=\frac{-dq}{p^2-4q}$\ $\implies(p^2-4q)(-dp)=(-4q)(-dq)$ $\implies(4q-p^2)(dp)=(4q)(dq)$ ON integrating both side we get $\implies4pq-\frac{-p^3}{3}=4pq+c$ "where c is a constant" $\implies(p^3)=c$

VIKAS MEENA
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