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When we solve quasi linear partial differential differential $Pp+Qq=R$ we make Lagrange’s equation like $\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$. Now my question is if $R=0$ then how we put $z=c$? Then we find second linearly independent solution by solving first two ration (say) $u$ and say general solution of the differential differential is $z=f(u)$. So what is reason behind to put $z$ as constant? Sorry for this silly question as I am new in differential equations. Thank you in advance.

Ymylife
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    Say you have a function $u = u(x(s), t(s))$, parameterised by $s$, that solves $Pu_{t} + Qu_{x} = R$. Then computing the total derivative of $u$ and equating to the PDE gives \begin{align} \frac{du}{ds} &= u_{t} t'(s) + u_{x} x'(s) \ &= P u_{t} + Q u_{x} \ &= R \end{align} Hence we must have $$t'(s) = P, \quad x'(s) = Q, \quad u'(s) = R$$ If $R \equiv 0$ then you are really solving the ODE $u'(s)= 0 \implies u(s) = \text{constant}$. The form you wrote the ODEs in $$\frac{dx}{P}=\frac{dy}{Q}=\frac{du}{R}$$ is the same as the ODEs I wrote down, its just the parameter $s$ has been eliminated. – Matthew Cassell Feb 05 '22 at 05:42
  • @mattos I think your comment is really the answer and you could put it as an answer, if you want to. – Bob Terrell Feb 05 '22 at 21:41

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