Question

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

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

Short answer

The solution of differential equation is$y=-\frac{\sqrt{x^2+y^2}}{x^2}$

Step-by-step solution

Given information

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

Definition of differential equation

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.

Solve the given differential equation

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

Replace the variable  in given differential equation

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 $\begin{array}{rcl}y& =& -\frac{\sqrt{x^2+y^2}}{x^2}\end{array}$.