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Integrate the differential equation xy′ = y – (x2 + y2)1/2

Integrate the following differential equation:

    \[ xy' = y - \sqrt{x^2 + y^2}. \]


Make the substitution y = vx giving y' = v + v'x. 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*}

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*}

But then, since

    \[ \left( v + \sqrt{1+v^2} \right)' = \left( 1 + \frac{v}{\sqrt{1+v^2}} \right) \]

this integral is of the from \int \frac{f'(v)}{f(v)} \, dv so we have

    \[ \int \frac{1}{\sqrt{1+v^2}} \, dv = \log \left( v + \sqrt{1+v^2} \right) + C. \]

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*}

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