Find the integral surface of tan x· p + y q = z, p = (∂…
Find the integral surface of \[ \tan x\cdot p + y\,q = z, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}, \] that passes through the curve $y=x$, $z=x^3$.
Explanation of Method
Because the initial data does not let $\Phi$ be solved for in closed elementary form, the resulting integral surface is naturally expressed in parametric form, with the parameter $t$ identifying the point of the initial curve each characteristic passes through.
Step 1 · Characteristic equations and first integrals
\[ \frac{dx}{\tan x} = \frac{dy}{y} = \frac{dz}{z}. \tag{1} \] From the last two ratios: $z/y=c_1$. \tag{I} From the first two: $\cot x\,dx=dy/y \Rightarrow y/\sin x=c_2$. \tag{II}
Step 2 · General solution
\[ z = y\,\Phi\!\left(\frac{y}{\sin x}\right). \tag{$*$} \]
Step 3 · Apply the initial condition (parametrically)
On $y=x$, $z=x^3$, parametrise by $t=x$, so along the characteristic through the initial point $t$: $c_1=z/y=t^2$ and $c_2=y/\sin x=t/\sin t$. Eliminating $t$ between $z=yc_1$ and $c_2=y/\sin x$ gives the surface implicitly in terms of the parameter $t$: \[ z = y t^2, \qquad y\sin t = t\sin x. \]
Verification
Setting $t=x$ recovers $y=x$ (from $y\sin x=x\sin x$) and then $z=yx^2=x^3$. $\checkmark$
Checking $c_1=z/y$ and $c_2=y/\sin x$ against $PF_x+QF_y+RF_z=0$ with $P=\tan x$,
$Q=y$, $R=z$ confirms both are first integrals:
\[ \tan x\cdot0+y\!\left(-\frac{z}{y^2}\right)+z\cdot\frac1y = 0, \qquad
\tan x\!\left(-\frac{y\cos x}{\sin^2x}\right)+\frac{y}{\sin x} = 0. \checkmark \]
Differentiating $z=y\Phi(y/\sin x)$ and substituting into $\tan x\cdot p+yq$
confirms it collapses to $y\Phi=z$. $\checkmark$
Result
The integral surface is given parametrically by: