Answer to Question #105581 in Differential Equations for Gayatri Yadav

Question

Find f(y) so that equation f(y)dx − zx dy − xy ln y dz = 0 is integrable. Also obtain the corresponding integral using Natani’s method.

Answer 1

Equation $f(y)dx-zxdy-xylnydz=0$ is integrable, if 

$X\cdot rotX=0\\ X=(f(y),-zx,-xylny)$

Calculate $rotX$ :

$rotX=\nabla\times X=\begin{vmatrix} \vec{i} & \vec{j}&\vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ f(y)&-zx&-xylny \end{vmatrix}=\\ =\vec{i}( \frac{\partial}{\partial y}(-xylny)- \frac{\partial}{\partial z}(-xz))-\\ -\vec{j}( \frac{\partial}{\partial x}(-xylny)- \frac{\partial}{\partial z}(f(y)))+\\ +\vec{k}( \frac{\partial}{\partial x}(-zx)- \frac{\partial}{\partial y}(f(y)))=\\ =(-xlny)\vec{i}+(ylny)\vec{j}+(-z-f'(y))\vec{k}$

Then

$X\cdot rotX=-xlny\cdot f(y)+xylnyf'(y)=0$

Thus $xlny\cdot(yf'(y)-f(y))=0$ , hence $yf'(y)-f(y)=0$

The solution of the last equation is determined by

$\frac{df}{f}=\frac{dy}{y}\\ ln|f(y)|=ln|y|+ln|A|\\ f(y)=A\cdot y$

where $A$ is a real constant. Finally, equation takes the form 

$Aydx-zxdy-xylnydz=0$

Using Natani’s method we firstly put $z=const$ and solve the obtained equation

$Aydx=zxdy\\ a\frac{dx}{x}=z\frac{dy}{y}\\ Alnx=zlny+F(z)$

Now we let $x=1$ . Then equation takes the form 

$-zdy-ylnydz=0$

and

$F(z)=-zlny$

Thus

$zdy+ylnydz=0\\ \frac{dy}{ylny}+\frac{dz}{z}=0\\ ln|lny|=-ln|z|+ln|C|\\ lny=\frac{C}{z}$

Therefore, 

$F(z)=-zlny=-z\cdot\frac{C}{z}=-C=K=const$

Finally,

$Alnx=zlny+K$

where $K$ is an arbitrary real constant.