Question

Differential equations are sometimes solved by having a clever idea. Here is a little exercise in cleverness: Although the differential equation $\left(x-\sqrt{x^2+y^2}\right)\mathit{d}\mathit{x}\mathbf{+}\mathit{y}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$is not exact, show how the rearrangement$\mathbf{(}\mathit{x}\mathit{d}\mathit{x}\mathbf{+}\mathit{y}\mathit{d}\mathit{y}\mathbf{)}\mathbf{/}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{=}\mathit{d}\mathit{x}$and the observation$\frac{\mathbf{1}}{\mathbf{2}}\mathit{d}\left(x^2+y^2\right)\mathbf{=}\mathit{x}\mathit{d}\mathit{x}\mathbf{+}\mathit{y}\mathit{d}\mathit{y}$can lead to a solution.

Short answer

It has been shown that$\sqrt{x^2+y^2}=x+c$ is a solution to the given differential equation.

Step-by-step solution

To Find the Differential Equation

Consider that $d\left(\sqrt{x^2+y^2}\right)=\frac{xdx}{\sqrt{x^2+y^2}}+\frac{ydy}{\sqrt{x^2+y^2}}......\left(1\right)$.

The differential equation $xdx+ydy=\sqrt{x^2+y^2}$, can be written as:

$\frac{xdx}{\sqrt{x^2+y^2}}+\frac{ydy}{\sqrt{x^2+y^2}}=dx......\left(2\right)$

Final Answer

Based on (1) and (2), we get.

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

The left side is the total differential of $\sqrt{x^2+y^2}$and the right side is the total differential of $x+c$. Thus, $\sqrt{x^2+y^2}=x+c$is a solution of the differential equation.

Hence, $\sqrt{x^2+y^2}=x+c$is a solution to the differential equation.