Question

(x^2+y^2)p+2xyq=dz/(x+y)z

Answer 1

The Langrage equation is:

$$(x^2+y^2)p+2xyq=(x+y)z$$

From the general equaton:

$$Pp + Qq = R$$

From the above equation,

$P = x^2+ y^2\\ Q = 2xy\\ R = (x+y)z$

From the auxilliary equation,

$$\dfrac{dx}{P} = \dfrac{dy}{Q} = \dfrac{dz}{R}$$

Hence:

$$\dfrac{dx}{x^2+y^2} = \dfrac{dy}{2xy} = \dfrac{dz}{(x+y)z} \qquad \cdots \cdots \cdots eqn(i)\\$$

From the first two and last ratios and solve, we have:

$$\dfrac{dx+dy}{(x^2+y^2)+2xy}=\dfrac{dz}{(x+y)z}\\$$

Recall:

$$x^2+y^2 = (x+y)^2 - 2xy \qquad \cdots eqn (*) \\$$ $$\therefore \dfrac{dx+dy}{(x+y)^2 -2xy +2xy}=\dfrac{\frac{dz}{z}}{x+y}\\ \dfrac{dx+dy}{(x+y)^2}=\dfrac{\frac{dz}{z}}{x+y}\\ \dfrac{dx+dy}{x+y}=\dfrac{dz}{z}\\ \dfrac{d(x+y)}{x+y}=\dfrac{dz}{z}\\$$

Integrating both sides:

$$\int \dfrac{d(x+y)}{x+y}=\int \dfrac{dz}{z}\\ \ln (x+y) = \ln z +\ln c\\ \ln (x+y) =\ln cz\\ x+y = cz\\ \implies c_1 = \dfrac{x+y}{z} \qquad \cdots\cdots eqn (ii)$$

Consider the last ratio in eqn (i):

$$\frac{dz}{(x+y)z}$$

From eqn (ii)

$$z = \dfrac{x+y}{c_1}$$

Substituting the value of $z$ in the last ratio:

$$\frac{dz}{(x+y)\frac{(x+y)}{c_1}}\\ \frac{dz}{\frac{(x+y)^2}{c_1}}\\ \implies \frac{c_1dz}{(x+y)^2} \qquad \cdots eqn (iii)$$

By replacing the third ration in eqn (i) with eqn(iii)

$$\frac{dx}{x^2+y^2} = \frac{dy}{2xy} = \frac{c_1dz}{(x+y)^2}$$

Thus:

$$\frac{dx-dy-c_1dz}{x^2+y^2-2xy- (x+y)^2} = k \qquad \text{(say)}$$

Using eqn (*):

$$\frac{dx-dy-c_1dz}{x^2+y^2-2xy- [x^2+y^2-2xy]} = k\\ \frac{dx-dy-c_1dz}{x^2+y^2-2xy- x^2-y^2+2xy} = k\\ \frac{dx-dy-c_1dz}{0} = k\\ \implies dx-dy-c_1dz = 0\\$$

Integrating through:

$$\int dx-\int dy-\int c_1dz = \int 0\\ x-y-c_1z=c_2\\$$

Recall:

$$c_1 = \dfrac{x+y}{z}$$

$$\therefore x-y-\Big(\frac{x+y}{z}\Big)z=c_2\\ x-y-x-y=c_2\\ \implies c_2 = -2y$$

Therefore, the solution is:

$$f(u,v) = \Big(\dfrac{x+y}{z},\; -2y\Big) = 0$$