# How to solve $\ln (v+\sqrt{1+v^2})=-k\ln(x)+k\ln(a)$ for $v$ in terms of $x$?

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

• Take sinh of both sides. en.wikipedia.org/wiki/… Sep 13 at 7:06
• $k ln(a) - k ln(x) = ln((a/x)^{k})$
– Lac
Sep 13 at 7:09

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 $$v$$.

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 $$x>0$$

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 $$v=\frac{1}{2}[(x/a)^k-(a/x)^k]$$