Use the methods to solve the following differential equations. Compare computer solutions and reconcile differences.

$\mathit{y}\mathit{d}\mathit{y}\mathbf{=}\left(-x+\sqrt{x^2+y^2}\right)\mathit{d}\mathit{x}$

The given differential equation is$ydy=\left(-x+\sqrt{x^2+y^2}\right)dx$

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself to its derivatives of various orders.

Homogenous equation:-

A homogeneous function of x and y of degree n means a function can be written as

$\begin{array}{rcl}& & x^{n}f\left(\frac{y}{x}\right)\\ P\left(x,y\right)dx+Q\left(x,y\right)dy& =& 0\end{array}$

Where P and Q are homogeneous function of the same degree is called homogeneous

By the change of variable of homogeneous equation

$\begin{array}{rcl}y\text{'}& =& \frac{dy}{dx}=-\frac{P\left(xy\right)}{\left(Cxy\right)}=f\left(\frac{y}{x}\right)\end{array}$

Replace by y = xv in equation

$\begin{array}{rcl}ydy& =& \left(-x+\sqrt{x^2+y^2}\right)dx\end{array}$Homogenous equation:-A homogeneous function of x and y of degree n means a function can be written as$\begin{array}{rcl}& & x^{n}f\left(\frac{y}{x}\right)\end{array}$

$\begin{array}{rcl}P(x,y)dx+Q(x,y)dy& =& 0\end{array}$

Where P and Q are homogeneous function of the same degree is called homogeneous

By the change of variable of homogeneous equation

$\begin{array}{rcl}y\text{'}& =& \frac{dy}{dx}\\ & =& -\frac{P\left(xy\right)}{\left(Qxy\right)}\\ & =& f\left(\frac{y}{x}\right)\end{array}$

Replace by y = xv in equation (1)

$\begin{array}{rcl}ydy& =& \left(-x+\sqrt{x^2+y^2}\right)dx\\ ydy& =& -xdx+\sqrt{x^2+y^2}dx\\ ydy+xdx-\sqrt{x^2+y^2}dx& =& 0\\ xy-\sqrt{x^2+{\left(vx\right)}^2}dx& =& 0\end{array}$

Solve further the equation

$\begin{array}{rcl}xy-\sqrt{x^2+x^2{v}^2}dx& =& 0\\ xy-x\sqrt{1+v^2}dx& =& 0\\ xy-\sqrt{1+v^2}& =& 0\\ y& =& \frac{\sqrt{1+v^2}}{x}\\ y& =& \frac{\sqrt{x^2+y^2}}{x^2}\end{array}$

Therefore, the solution of differential equation is $\begin{array}{rcl}y& =& -\frac{\sqrt{x^2+y^2}}{x^2}\end{array}$.