Answer to Question #115575 in Differential Equations for Neha

Question

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

By Lagrange’s method the auxiliary equations of the given differential equation are as following:

$\frac{𝑑𝑥}{𝑥−𝑦}=\frac{𝑑𝑦}{𝑦−𝑥−𝑧}=\frac{𝑑𝑧}{𝑧}$

$\implies \frac{𝑑𝑥}{𝑥−𝑦}=\frac{𝑑𝑦}{𝑦−𝑥−𝑧}=\frac{𝑑𝑧}{𝑧} = \frac{dx+dy+dz}{0} \\ \implies dx+dy+dz =0 \implies x+y+z = c_1$

Hence from the first fraction, we have $\frac{𝑑𝑥}{𝑥−𝑦}=\frac{𝑑𝑦}{𝑦−𝑥−𝑧}$

$\implies \frac{𝑑𝑦}{𝑦−c_1+y}=\frac{𝑑z}{z}$

Now, by integrating on both sides we get $(1/2)\log(2y-c_1) =\log(z)+\log(\sqrt{c_2}) \implies c_2 = \frac{y-x-z}{z^2}$ .

Since $x+y+z=c_1$ and $\frac{y-x-z}{z^2}=c_2$ are independent solutions of the given differential equation, so a solution of the given differential equation is

$c_2= f(c_1) \implies \frac{y-x-z}{z^2} = f(x+y+z)$ ________(1)

On $x^2+y^2 = 1, z=1$ , so put $x= \cos{(t)}, y = \sin{(t)}$ and put in solution, we get

$\sin{(t)}-\cos{(t)}=f(\cos{(t)}+\sin{(t)})$ ________(2)

So, the solution of the given differential equation is given by (1), where the function is given by (2).