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Question The given partial differential equation is $$px+qy+pq=xz$$ My approach Where $p=\frac{\partial z}{\partial x}$ and $q= \frac{\partial z}{\partial y}$

By using charpit method $$ \frac{dx}{F_p}=\frac{dy}{F_q}=\frac{dz}{pF_p+qF_q} =\frac{-dp}{F_x+pF_z}=\frac{-dq}{F_y+qF_z} $$ Putting the values

$$\frac{dx}{x+q}=\frac{dy}{xy+p}=\frac{dz}{px+qxy+2pq}=\frac{-dp}{p+qy-z-px}=\frac{-dq}{qx-qx}$$ From the last fraction of $dq$ we get $q=a$, where $a$ is arbitary constant.

Substituting the value of $q$ in given PDE we get $p$ as $$p= \frac{x(z-ay)}{x+a}$$ Further the solution of the differential equation be $dz=pdx+qdy$.

From above we get $$dz= \left[\frac{x(z-ay)}{x+a} \right]dx + ady $$ My question

1) How can i integrate the above formed equation, as i am unable to do so.

2) Is there any chance of converting above given equation into standard clairaut's form ?

Any subtle hint is highly appreciated Thankyou

Answer 1

If the PDE is $$px+qy+pq=xz\:,$$ the Charpit's equations are : $$\frac{dx}{x+q}=\frac{dy}{y+p}=\frac{dz}{px+qy+2pq}=\frac{-dp}{p-z-px}=\frac{-dq}{q-qx}$$ This is not what you wrote.

Then $z=ay+be^x(x+a)^{-a}$ is not solution of the above PDE. $$ $$

If the Charpit's equations are : $$\frac{dx}{x+q}=\frac{dy}{xy+p}=\frac{dz}{px+qxy+2pq}=\frac{-dp}{p+qy-z-px}=\frac{-dq}{qx-qx}$$ The PDE is : $$px+qxy+pq=xz$$ This is not the PDE written in your question.

$$\boxed{z=ay+be^x(x+a)^{-a}\quad\text{is solution of}\quad px+qxy+pq=xz}$$

Answer 2

From the last equation formed in question $$dz = \left[\frac{x(z-ay)}{x+a}\right]dx + ady $$ $$\frac{dz}{z-ay}= \frac{x dx}{x+a} + \frac{a dy}{z-ay} $$ And $$\frac{dz-ady}{z-ay} = \frac{x+a}{x+a} + \frac{a}{x+a} $$ Integrating the above equation $$ log_e(z-ay) = x -alog_e(x+a) + log_e{b}$$ Where $log_e{b}$ is arbitrary constant

$$ log_e(z-ay) = x+ log_e(x+a)^{-a} +log_e{b} $$ And the final answer be $$z = be^{x}(x+a)^{-a} +ay$$

edit :- added the missing term $ay$ in Right hand side. As suggested by JJacquelin