Question

Find the general integral of the equation ( x - y )p + (y - x - z)q = z and the particular solution through the circle z = 1, x2 + y2 = 1 .

13 Feb 2018
Answer :
Word Count : 283
Consider the first-order linear PDE with $p=\partial z/\partial x,\; q=\partial z/\partial y$: $$ (x-y)p+(y-x-z)q=z. $$ Form the characteristic system $$ \frac{dx}{x-y}=\frac{dy}{y-x-z}=\frac{dz}{z}. $$ From $\dfrac{dx}{x-y}=\dfrac{dz}{z}$ and $\dfrac{dy}{y-x-z}=\dfrac{dz}{z}$ we may use a combination: compute $d(x+y)/ds$ where $s$ is the characteristic parameter. Since $$ \frac{dx}{ds}=x-y,\qquad \frac{dy}{ds}=y-x-z,\qquad \frac{dz}{ds}=z, $$ we have $$ \frac{d}{ds}(x+y)=(x-y)+(y-x-z)=-z, \qquad \frac{dz}{ds}=z. $$ Hence $$ \frac{d(x+y)}{dz}=\frac{-z}{z}=-1, $$ so integrating gives the first integral (constant along characteristics) $$ x+y+z=C_1. $$ Use this constant $C_1$ to reduce the system. From $x+y+z=C_1$ we have $y=C_1-x-z$. Substitute into $\dfrac{dx}{dz}=\dfrac{x-y}{z}$: $$ \frac{dx}{dz}=\frac{x-(C_1-x-z)}{z}=\frac{2x+z-C_1}{z}=\frac{2x}{z}+1-\frac{C_1}{z}. $$ This is a linear ODE in $x(z)$: $$ \frac{dx}{dz}-\frac{2}{z}x=1-\frac{C_1}{z}. $$ The integrating factor is $z^{-2}$.
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