Question

yz(y+z)dx+zx(z+x)dy+xy(x+y)dz=0

Answer 1

Given differential equation is

$yz(y+z)dx +xz(x+z)dy+xy(x+y)dz = 0$

Let $\vec{X} = yz(y+z)\hat{i} +xz(x+z)\hat{j}+xy(x+y)\hat{k}$

To check whether it is integral , $\vec{X} .(\nabla\times{\vec{X}}) = 0$

Now, $\nabla\times{\vec{X}} = \begin{vmatrix} \hat{i} & \hat{j} &\hat{k}\\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} &\frac{\partial }{\partial z}\\ yz(y+z) & xz(x+z) & xy(x+y) \end{vmatrix}$

$\nabla\times{\vec{X}} = 2x(y-z )\hat{i} - 2y(x-z)\hat{j} + 2z(x-y)\hat{k}$

Now, $\vec{X} .(\nabla\times{\vec{X}}) =[ yz(y+z)\hat{i} +xz(x+z)\hat{j}+xy(x+y)\hat{k} ] .[2x(y-z )\hat{i} - 2y(x-z)\hat{j} + 2z(x-y)\hat{k}] = 0$

Hence it it intergrable.

Let $y = ux \implies dy = xdu + udx$

and $z = vx \implies dz = vdx+xdv$

Then partial differential equation will be reduced to the form

$uxvx(ux+vx)dx + vx^2(x+vx)(udx+xdu) + ux^2(x+ux)(vdx+xdv) = 0$

It can be further reduced to

$\frac{dx}{x} + \frac{du}{2u}+\frac{dv}{2v}-\frac{du}{2(u+v+1)}-\frac{dv}{2(u+v+1)} = 0$

Integrating it, we get

$\frac{x^2 uv}{u+v+1} = C$

putting values of u and v,

$xyz = C(x+y+z)$