Answer to Question #238505 in Differential Equations for vaibhav

Question

Find the integral surface of 𝑥²𝑝 + 𝑦²𝑞 + 𝑧² = 0 which passes through the curve x+y = xy, 𝑧= 1.

Answer 1

x²p+y²q+z²=0

x²p+y²q=-z²

By lagrange equation;

$\frac{dx}{x^{2}}=\frac{dy}{y^{2}}=\frac{dz}{-z^{2}}$

Taking the 1^st and 2^nd terms,

$\frac{dx}{x^{2}}=\frac{dy}{y^{2}}$

Integrate both sides,

$\int \frac{dx}{x^{2}}=\int \frac{dy}{y^{2}}$

$C_1+\frac{1}{-x}=-\frac{1}{y}$

$C_1=\frac{1}{x}-\frac{1}{y}$ This is equation (i)

Taking the 2^nd and 3^rd terms,

$\frac{dy}{y^{2}}=\frac{dz}{-z^{2}}$

Integrate both sides

$\int \frac{dy}{y^{2}}=\int \frac{dz}{-z^{2}}$

$-\frac{1}{y}+C_2=\frac{1}{z}$

$C_2=\frac{1}{y}+\frac{1}{z}$ This is equation (ii)

The equation of the curve given is x+y=xy and z=1

We let x=t

Therefore from the equation of the curve;

x=xy-y

Rearrange; xy-y=x

y(x-1)=x

$y=\frac{x}{x-1}\implies y=\frac{t}{t-1}$

z=1

Put the above values of x,y and z in equations (i) and (ii) and remove t from the obtained relation

$C_1=\frac{1}{t}-\frac{t-1}{t}$

$C_1=\frac{1}{t}-1+\frac{1}{t}$

$C_1=\frac{1}{t}+\frac{1}{t}-1$ This is equation (iii)

$C_2=1-\frac{1}{t}+1$

$C_2=2-\frac{1}{t}$

$\frac{1}{t}=2-C_2$ Replace this in equation (iii)

$C_1=2-C_2+2-C_2-1$

$C_1=3-2C_2$ This is equation (iv)

Now replace the values of C₁ and C₂ in (iv) to get the required surface

$\frac{1}{x}-\frac{1}{y}=3-\frac{2}{y}-\frac{2}{z}$

$\frac{y-x}{xy}=\frac{3yz-2z-2y}{yz}$

$zy-xz=3xyz-2xz-2xy$

The Integral surface is;

$zy+xz+2xy=3xyz$