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Consider the differential equation

$$\frac{dy}{dx}=\frac{y}{x}-k\sqrt{1+\frac{y^2}{x^2}}$$

We can utilize the substitution $v=\frac{y}{x}$ to solve this equation for $y(x)$.

$$\frac{dy}{dx}=\frac{dv}{dx}x+v=v-k\sqrt{1+v^2}$$ $$\frac{1}{\sqrt{1+v^2}}dv=-\frac{k}{x}dx$$ $$ln (v+\sqrt{1+v^2})=-kln(x)+C$$

Assume $y(a)=0$ is the initial condition. Then $v(a)=\frac{y(a)}{x}=0$ and $C=kln(a)$ $$ln (v+\sqrt{1+v^2})=-kln(x)+kln(a)$$

We need to solve this for $v$ in terms of x, then we can substitute in $v=\frac{y}{x}$ to obtain $y(x)$, the solution the original differential equation.

The question is, how does one solve for $v$ here?

This is all from an example in a differential equations textbook. To be exact, it is chapter 1.6 (Substitution Methods and Exact Equations) of Edwards/Penney's Elementary Differential Equations, 6th edition. The differential equation here is from a problem on calculating a flight trajectory. The example doesn't show the passage that this question is about; it is rather given as an end-of-chapter problem, but I don't have solutions

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Using @Lac's comment, we have $$\ln(v+\sqrt{1+v^2})=\ln\left(\left(\frac{a}{x}\right)^k\right)$$ $$\Rightarrow v+\sqrt{1+v^2}=\left(\frac{a}{x}\right)^k \tag{1}$$

From the identity $a^2-b^2=(a-b)(a+b)$, we have $$\Rightarrow a-b=\frac{a^2-b^2}{a+b}$$ where $a\neq b$.

So, we can set $a=\sqrt{1+v^2}, b=v$ and obtain $$\sqrt{1+v^2}-v=\frac{\left(\sqrt{1+v^2}\right)^2-v^2}{\sqrt{1+v^2}+v}$$ $$\Rightarrow \sqrt{1+v^2}-v=\left(\frac xa\right)^k \tag{2}$$

Now you can add (or subtract) $(1)$ and $(2)$ and solve for $v$.

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The better antiderivative to use is

$$\frac{dv}{\sqrt{1+v^2}} = -\frac{k}{x}dx \implies \sinh^{-1}v = -k \ln |x| + C$$

which means

$$y = x \sinh \left(k\ln\left|\frac{a}{x}\right|\right)$$

or

$$y = \frac{a^k}{2}x^{1-k}-\frac{a^{-k}}{2}x^{1+k}$$

for $x>0$

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It is a simple algebraic simplification. $$\log[\sqrt{1+v^2}+v]=\log (a/x)^k$$ $$\implies [\sqrt{1+v^2}+v]=(x/a)^k$$ $$\implies [\sqrt{1+v^2}-v]=(a/x)^k$$ Subtract the two to get $$v=\frac{1}{2}[(x/a)^k-(a/x)^k]$$

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