Integrate the following differential equation:
![\[ xy' = y - \sqrt{x^2 + y^2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1c7a1a657934e1474f7cfe00469ff01d_l3.png "Rendered by QuickLaTeX.com")
Make the substitution 
giving 
. Then we have,
![\begin{align*} && xy' &= y - \sqrt{x^2+y^2} \\[9pt] \implies && y' &= \frac{y}{x} - \sqrt{1 + \left( \frac{y}{x} \right)^2} \\[9pt] \implies && v + v'x &= v - \sqrt{1+v^2} \\[9pt] \implies && \frac{1}{\sqrt{1+v^2}} v' &= -\frac{1}{x} \\[9pt] \implies && \int \frac{1}{\sqrt{1+v^2}} \, dv &= - \int \frac{1}{x} \, dx. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-40eed83bcf95e355ea44cccb4a57d4d9_l3.png "Rendered by QuickLaTeX.com")
Evaluating the integral on the left is a bit tricky (and I can’t find an exercise where we’ve already done it). For this integral we have,
![\begin{align*} \int \frac{1}{\sqrt{1+v^2}} \, dv &= \int \frac{v + \sqrt{1+v^2}}{\sqrt{1+v^2} \cdot (v + \sqrt{1+v^2})} \, dv \\[9pt] &= \int \left( \frac{1}{v + \sqrt{1+v^2}} \cdot \frac{\sqrt{1+v^2} + v}{\sqrt{1+v^2}} \right) \, dv \\[9pt] &= \int \left( \frac{1}{v+ \sqrt{1+v^2}} \cdot \left( 1 + \frac{v}{\sqrt{1+v^2}} \right) \right) \, dv. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0dcfc55aebdb694879d359db3e9c38f0_l3.png "Rendered by QuickLaTeX.com")
But then, since
![\[ \left( v + \sqrt{1+v^2} \right)' = \left( 1 + \frac{v}{\sqrt{1+v^2}} \right) \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a219ab02e0c7567a848a4aa69d35aa01_l3.png "Rendered by QuickLaTeX.com")
this integral is of the from 
so we have
![\[ \int \frac{1}{\sqrt{1+v^2}} \, dv = \log \left( v + \sqrt{1+v^2} \right) + C. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-203b93335935af30811cafe064c5f645_l3.png "Rendered by QuickLaTeX.com")
Putting this back into where we left on our differential equation, we then have
![\begin{align*} &&\int \frac{1}{\sqrt{1+v^2}} \, dv &= - \int \frac{1}{x} \, dx \\[9pt] \implies && \log \left( v + \sqrt{1+v^2} \right) &= - \log |x| + C \\[9pt] \implies && v + \sqrt{1+v^2} &= \frac{C}{x} \\[9pt] \implies && \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} &= \frac{C}{x} \\[9pt] \implies && y + \sqrt{x^2 + y^2} &= C \\[9pt] \implies && x^2 + y^2 &= (C-y)^2 \\[9pt] \implies && x^2 + y^2 &= C^2 - 2Cy + y^2 \\[9pt] \implies && x^2 + 2Cy &= C^2. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-834f04c4fc51d7dcf11431606ad525d2_l3.png "Rendered by QuickLaTeX.com")
This integral can also be evaluated as arg sh (v) , and then using Euler relation to sh ( log(Cx)) getting 2y=1/C-Cx^2, that is also x^2+2Cy=C^2, with C>0.
This integral can be evaluated without the trick by substituting tan(t) instead of v, and then doing simple derivatives involving sec(t)