Question

Solve (x+y)(p+q) =z-1

Answer 1

Given $(x+y)(p+q) =z-1$

$\implies (x+y)p + (x+y)q = z-1$

So, Lagrange's equation is

$\frac{dx}{x+y} = \frac{dy}{x+y} = \frac{dz}{z-1}$ .

By First two factors of Lagrange's equation

$\frac{dx}{x+y} = \frac{dy}{x+y} \\ \implies dx = dy \\ \implies x - y =c_1$ ______________(1)

By last two factors, we get

$\frac{dy}{x+y} = \frac{dz}{z-1}$

Now, by using solution (1), we get

$\frac{dy}{c_1+2y} = \frac{dz}{z-1}$

By integrating, we get

$\frac{1}{2}\log(c_1+2y) = \log(z-1)+c_2$

$\implies \frac{1}{2}\log(x+y) = \log(z-1)+c_2$

$\implies 0 = \log(\frac{z-1}{\sqrt{x+y}})+c_2$

$\implies \frac{z-1}{\sqrt{x+y}} = e^{-c_2} = c_3$ ____________(2)

Hence, Solution is $c_3 = f(c_1)$

$\implies z-1 = (\sqrt{x+y}) \ f(x-y)$