Question

a) Find the differential equation of the space curve in which the two families of
surfaces
u = x^2 − y^2 = c1 and v=y^2−z^2=c2 intersect.
b) Find the general integral of the equation
( x − y)p + (y − x − z) q = z
and a particular solution through the circle z=1,x^2+y^2=1

Answer 1

SOLUTION (a)

Given two families of surfaces  

$$u=x^2-y^2=c_1\quad\text{and}\quad v=y^2-z^2=c_2$$

For all level surface

$$du=0\longrightarrow du=2xdx-2ydy=0\longrightarrow\\[0.3cm] \boxed{xdx=ydy}\\[0.3cm] dv=0\longrightarrow dv=2ydy-2zdz=0\longrightarrow\\[0.3cm] \boxed{ydy=zdz}\\[0.3cm]$$

Then,

$$\left.xdx=ydy=zdz\right|\div(xyz)\\[0.3cm] \frac{xdx}{xyz}=\frac{ydy}{xyz}=\frac{zdz}{xyz}\\[0.3cm] \frac{dx}{yz}=\frac{dy}{xz}=\frac{dz}{xy}\,\text{is auxiliary equation}\\[0.3cm]$$

Conclusion,

$$\boxed{(yz)p+(xz)q=xy\,\text{is desired equation}}$$

SOLUTION (b)

Given the differential equation

$$(x-y)p+(y-x-z)q=z$$

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

$$\frac{dx}{(x-y)}=\frac{dy}{y-(x+z)}=\frac{dz}{z}$$

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

$$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}=\ldots=\frac{\lambda_1a_1+\lambda_2a_2+\cdots+\lambda_na_n}{\lambda_1b_1+\lambda_2b_2+\cdots+\lambda_nb_n}$$

In our case, Choosing $\lambda_1=\lambda_2=\lambda_3=1$ as multipliers, each fraction on

$$\frac{dx}{(x-y)}=\frac{dy}{y-(x+z)}=\frac{dz}{z}=\frac{1\cdot dx+1\cdot dy+1\cdot dz}{1\cdot(x-y)+1\cdot(y-x-z)+z}\\[0.3cm] \frac{dx}{(x-y)}=\frac{dy}{y-(x+z)}=\frac{dz}{z}=\frac{dx+dy+dz}{0}\longrightarrow\\[0.3cm] d(x+y+z)=0\longrightarrow\boxed{x+y+z=C_1}$$

Take the last two fractions of auxiliary equation and using $x+y+z=C_1$ we get

$$\left\{\begin{array}{l} \displaystyle\frac{dy}{y-(x+z)}=\displaystyle\frac{dz}{z}\\[0.3cm] x+y+z=C_1 \end{array}\right.\longrightarrow\left\{\begin{array}{l} \displaystyle\frac{dy}{y-(C_1-y)}=\displaystyle\frac{dz}{z}\\[0.3cm] (x+z)=C_1-y \end{array}\right.\\[0.3cm] \int\frac{dy}{2y-C_1}=\int\frac{dz}{z}\longrightarrow\\[0.3cm] \frac{\ln|2y-C_1|}{2}=\ln|z|+\ln|C_2|\longrightarrow\ln|2y-C_1|-2\ln|z|=2\ln|C_2|\\[0.3cm] \ln\left|\frac{2y-C_1}{z^2}\right|=\ln|\underbrace{C_2^2}_{C_3}|\longrightarrow\frac{2y-(x+y+z)}{z^2}=C_3\\[0.3cm] \boxed{\frac{y-x-z}{z^2}=C_3}$$

We have found two integrals for the given equation

$$\left\{\begin{array}{l} x+y+z=C_1\\[0.3cm] \displaystyle\frac{y-x-z}{z^2}=C_3 \end{array}\right.$$

Therefore, any integral surface of the differential equation $(𝑥−𝑦)𝑝+(𝑦−𝑥−𝑧)𝑞=𝑧$ is described by the equation

$$\varphi\left(C_1,C_3\right)=0\longrightarrow\\[0.3cm] \boxed{\varphi\left((x+y+z),\frac{y-x-z}{z^2}\right)=0\,\text{is the general integral}}$$

where $\varphi$ is an arbitrary smooth function.

Particular solution:

A particular solution through the circle $z=1, x^2+y^2=1$

$$\left\{\begin{array}{l} x+y+z=C_1\\[0.3cm] \displaystyle\frac{y-x-z}{z^2}=C_3\\[0.3cm] z=1\\[0.3cm] x^2+y^2=1 \end{array}\right.\longrightarrow \left\{\begin{array}{l} x+y+1=C_1\\[0.3cm] y-x-1=C_3 \end{array}\right.\\[0.3cm] \left\{\begin{array}{l} x+y+1=C_1\\[0.3cm] y-x-1=C_3 \end{array}\right.\longrightarrow\left\{\begin{array}{l} y=\displaystyle\frac{C_1+C_3}{2}\\[0.3cm] x=\displaystyle\frac{C_1-C_3}{2}-1 \end{array}\right.\\[0.3cm]$$

Substituting the found expressions into the second condition

$$x^2+y^2=1\longrightarrow\left(\frac{C_1-C_3}{2}-1\right)^2+\left(\frac{C_1+C_3}{2}\right)^2=1\\[0.4cm] \frac{C_1^2-2C_1C_3+C_3^2}{4}-(C_1-C_3)+1+\frac{C_1^2+2C_1C_3+C_3^2}{4}=1\\[0.4cm] \frac{2C_1^2+2C_3^2}{4}-C_1+C_3=0\\[0.4cm] C_1^2+C_3^2-2C_1+2C_3=0\\[0.4cm] \boxed{C_1(C_1-2)+C_3(C_3+2)=0}$$

Since, $\varphi(C_1,C_3)=0$ , so

$$\varphi(C_1,C_3)=C_1(C_1-2)+C_3(C_3+2)-\text{Particular solution}$$

Conclusion,

$$\left\{\begin{array}{l} x+y+z=C_1\\[0.3cm] \displaystyle\frac{y-x-z}{z^2}=C_3 \end{array}\right.\longrightarrow\\[0.3cm] (x+y+z)(x+y+z-2)+\frac{y-x-z}{z^2}\cdot\left(\frac{y-x-z}{z^2}+2\right)=0\\[0.3cm]$$

ANSWER

$$\varphi\left((x+y+z),\frac{y-x-z}{z^2}\right)=0\,\text{is the general solution}\\[0.3cm] \text{Particular solution is}\\[0.3cm] (x+y+z)(x+y+z-2)+\frac{y-x-z}{z^2}\cdot\left(\frac{y-x-z}{z^2}+2\right)=0$$