Question

Find the integral surface of the partial differential equation:(x–y) y^2 p+(y–x)x^2 q=(x^2 +y^2)z through the curve xz=a^2, y=0

Answer 1

$(x-y)y^2p + (y-x)x^2q=(x^2+y^2)z$ is the equation to be solved for the curves

$xz=a^2$ ---(1)

$y=0$ ---(2)

Lagrange's auxillary equation is :

$dx/((x-y)y^2) = dy/((y-x)x^2)=dz/((x^2+y^2)z)$

Solving the first two sides; we get;

$\int (dx/x^2)= -\int (dy/y^2)$

$\implies x^3+y^3=c_1$ ---(3)

Now using $-1/(x-y),1/(x-y);1/z$ as Lagrange multipliers, we get;

$-dx/(x-y)+dy/(x-y)+dz/z=0$

$\implies dz/z=dx-dy/(x-y)$

Let $x-y=t \implies dx-dy=dt$

$\therefore \int dz/z= \int dt/t$

On solving we get;

$t/z=c_2 \implies (x-y)/z=c_2$ ---(4)

Using (2) in (3) and (4) we get;

$x_3=c_1$ ---(5)

$x=c_2z$

Now using (1) to substitute for z in the above equation, we get;

$x=ac^{1/2}$ ---(6)

Using (6) in (5), we get;

$c_1^2=c_2^3a^6$ ---(7)

Now substituting $c_1, c_2$ in (7) from (3) and (4), we get;

$(x^3+y^3)^2=(x-y)^3a^6/z^3$ ----(Answer)