Find the general solution of e^x p + y^2 q = y e^x, p = (∂…

Find the general solution of \[ e^x p + y^2 q = y e^x, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]

Step 1 · Characteristic equations

\[ \frac{dx}{e^x} = \frac{dy}{y^2} = \frac{dz}{ye^x}. \tag{1} \]

Step 2 · First integral

From $\dfrac{dx}{e^x}=\dfrac{dy}{y^2}$: $e^{-x}dx = dy/y^2$. Integrating: \[ -e^{-x} = -\frac{1}{y}+\text{const} \;\Longrightarrow\; \frac{1}{y}-e^{-x}=c_1. \tag{I} \]

Step 3 · Second integral

From $\dfrac{dy}{y^2}=\dfrac{dz}{ye^x}$: $e^x\,dy=y\,dz \Rightarrow \dfrac{dy}{y}=\dfrac{dz}{e^x}$. Along a characteristic, from (I), $e^{-x}=1/y-c_1$, so $e^x=\dfrac{y}{1-c_1y}$. Substituting: \[ \frac{dy}{y} = \frac{1-c_1y}{y}\,dz \;\Longrightarrow\; dz = \frac{dy}{1-c_1y}. \] Integrating: $z=-\dfrac{1}{c_1}\ln(1-c_1y)+c_2$. Since $1-c_1y = 1-\left(\tfrac1y-e^{-x}\right)y = ye^{-x}$, $\ln(1-c_1y)=\ln y - x$, so \[ c_2 = z + \frac{(\ln y - x)}{\tfrac1y-e^{-x}} = z - \frac{(x-\ln y)\,ye^x}{e^x-y}. \tag{II} \]

Step 4 · General solution

The general solution is $c_2=\Phi(c_1)$: \[ z = \frac{(x-\ln y)\,ye^x}{e^x - y} + \Phi\!\left(\frac{1}{y} - e^{-x}\right), \] where $\Phi$ is an arbitrary differentiable function.

Verification of the first integral

Along a characteristic parameterised by $t$ with $\dot x=e^x$, $\dot y=y^2$: \[ \dot c_1 = -\frac{\dot y}{y^2}+e^{-x}\dot x = -1+1 = 0. \checkmark \] The construction of the second integral (II) directly inverts the characteristic ODE obtained from the second and third ratios, so by construction the boxed relation satisfies the original PDE.