Question

Find the equation of the integral surface of the differential equation
(x^2 − yz) p + ( y^2 − zx) q = z^2 − xy
which passes through the line x =1, y = 0

Answer 1

Answer on Question 65076 - Math - Differential Equations

Question: Find the equation of the integral surface of the differential equation $(x^{2}-yz)p+(y^{2}-zx)q=z^{2}-xy$ which passes through the line $x=1$, $y=0$.

Solution: Consider a quasilinear equation

$a(x,y,z)p+b(x,y,z)q=c(x,y,z).$

By Lagrange’s method the auxiliary equations are as following:

$\frac{dx}{a(x,y,z)}=\frac{dy}{b(x,y,z)}=\frac{dz}{c(x,y,z)}.$

So, for the given quasilinear equation we come to the system in the symmetric form

$\frac{dx}{x^{2}-yz}=\frac{dy}{y^{2}-zx}=\frac{dz}{z^{2}-xy}.$ (1)

One of a way to solve the system in symmetric form is to use the equal fractions property

$\frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=\cdots=\frac{a_{n}}{b_{n}}=\frac{\lambda_{1}a_{1}+\lambda_{2}a_{2}+\cdots+\lambda_{n}a_{n}}{\lambda_{1}b_{1}+\lambda_{2}b_{2}+\cdots+\lambda_{n}b_{n}}.$

In our case we have

$\frac{dx-dy}{x^{2}-yz-y^{2}+zx}=\frac{dy-dz}{y^{2}-zx-z^{2}+xy},$

$\frac{d(x-y)}{x^{2}-y^{2}+z(x-y)}=\frac{d(y-z)}{y^{2}-z^{2}+x(y-z)},$

$\frac{d(x-y)}{(x-y)(x+y+z)}=\frac{d(y-z)}{(y-z)(x+y+z)},$

$\frac{d(x-y)}{x-y}=\frac{d(y-z)}{y-z}.$

Integrating of the last equality yields

$\int\frac{d(x-y)}{x-y}=\int\frac{d(y-z)}{y-z}$

$\Rightarrow\quad\ln|x-y|=\ln|y-z|+\ln|C_{1}|\quad\Rightarrow\quad\frac{x-y}{y-z}=C_{1}.$

Since the system (1) is invariant under the cyclic interchange of variables $x\mapsto y\mapsto z\mapsto x$, there exist two integrals

$\frac{x-y}{y-z}=C_{1},\qquad\frac{y-z}{z-x}=C_{2}.$

Therefore any integral surface of the differential equation $(x^{2} - yz)p + (y^{2} - zx)q = z^{2} - xy$ is described by the equation

$$F \left(\frac {x - y}{y - z}, \frac {y - z}{z - x}\right) = 0,$$

where $F$ is a smooth function. If we substitute the conditions $x = 1$ and $y = 0$, then

$$F \left(- \frac {1}{z}, \frac {- z}{z - 1}\right) = 0 \text{ or } F \left(\tau , - \frac {1}{\tau + 1}\right) = 0.$$

We see that the last relation does not define the function $F$ uniquely. Hence, the problem admits many solutions.

**Answer:** $F\left(\frac{x - y}{y - z}, \frac{y - z}{z - x}\right) = 0$, where $F = F(\xi, \eta)$ is a $C^1$-function such that $F\left(\tau, -\frac{1}{\tau + 1}\right) = 0$.

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