I am currently trying to find the curve that a surfer makes in attempting to get to beyond where the waves break while a current pushes against him perpendicularly, and am using this post as help: Name that curve!
I've gotten to a certain point but Wolfram Alpha doesn't explain the steps well, and I'm stuck.
I'm trying to go from this differential equation
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y – \rho r}{x},\quad r=\sqrt{x^2+y^2}$$
to this equation for $y$:
$$y(x) = x \sinh(\sinh^{-1}(y_0) – \rho \log x)$$
WolframAlpha succeeds in getting from the first to the second but I don't understand how it integrates the left side at one point, and where the $\sinh$ comes from…
WolframAlpha's approach
Step 1: Solve $\displaystyle \frac{\mathrm{d}y(x)}{\mathrm{d}x} = \frac{- \rho \sqrt{x^2 + (y(x))^2} + y(x)}{x}$, such that $y(1) = y_0$:
Step 2: Let $y(x) = x v(x)$, which gives $\displaystyle \frac{\mathrm{d}y(x)}{\mathrm{d}x} = x \: \frac{\mathrm{d}v(x)}{\mathrm{d}x} + v(x)$:
$$x \: \frac{\mathrm{d}v(x)}{\mathrm{d}x} + v(x) = \frac{- \rho \sqrt{x^2 + x^2 (v(x))^2} + x \hspace{0.25mm} v(x)}{x}$$Step 3: Simplify:
$$x \hspace{0.5mm} \frac{\mathrm{d}v(x)}{\mathrm{d}x} + v(x) = -\rho \sqrt{(v(x))^2 + 1} + v(x)$$Step 4: Solve for $\displaystyle \frac{\mathrm{d}v(x)}{\mathrm{d}x}:$
$$ \frac{\mathrm{d}v(x)}{\mathrm{d}x} = -\frac{\rho \sqrt{(v(x))^2 + 1}}{x} $$Step 5: Divide both sides by $\sqrt{(v(x))^2 + 1}$:
$$ \frac{\frac{\mathrm{d}v(x)}{\mathrm{d}x}}{\sqrt{(v(x))^2 + 1}} = – \frac{\rho}{x} $$Step 6 Integrate both sides with respect to $x$:
$$ {\int} \frac{\frac{\mathrm{d}v(x)}{\mathrm{d}x}}{\sqrt{(v(x))^2 + 1}} \mathrm{d}x = \int – \frac{\rho}{x} \mathrm{d}x $$Step 7: Evaluate the integrals:
$$\sinh^{-1}(v(x)) = – \rho \log(x) + c_1,$$ where $c_1$ is an arbitrary constant.Step 8: Solve for $v(x)$:
$$v(x) = -\sinh(\rho \log(x) – c_1).$$Step 9: Simplify the arbitrary constants:
$$v(x) = -\sinh(\rho \log(x) + c_1)$$Step 10: Substitute back for $y(x) = x \hspace{0.25mm} v(x)$:
$$y(x) = -x \sinh(\rho \log(x) + c_1)$$Step 11: Solve for $c_1$ using the initial conditions:
Substitute $y(1) = y_0$ into $y(x) = -x \sinh(\rho \log(x) + c_1):$
$$-\sinh(c_1) = y_0$$Step 12: Solve the equation: $c_1 = -\sinh^{-1}(y_0)$
Step 13: Substitute $c_1 = -\sinh^{-1}(y_0)$ into $y(x) = -x \sinh(\rho \log(x) + c_1)$:
Answer:
$$\boxed{y(x) = x \sinh(\sinh^{-1}(y_0) – \rho \log x)}$$
I understand steps 1-5, but I don't understand how to continue from Step 6. Steps 7-11 seem to just randomly come out.
If someone can explain in a more human friendly way than Wolfram Alpha what is going on here I would be extremely appreciative…
Thank you!
Best Answer
Essentially this is the substitution rule. More formally, the substitution $V=v(x')$ yields $$\int_0^{x}\frac{dv/dx'}{\sqrt{1+v(x')^2}}\,dx'=\int_0^{v(x)}\frac{1}{\sqrt{1+V^2}}dV$$ where I've switched to definite integration to make sure the changes of variable are all clear. Since $$\frac{d}{dV}\sinh^{-1} V=\frac{1}{\sqrt{1+V^2}},$$ this integrates to $ \Big[\sinh^{-1}V\Big]_{V=0}^{v(x)}=\sinh^{-1} v(x).$ In solutions, the above definite is instead typically presented as the indefinite integral $$\int \frac{dv/dx}{\sqrt{1+v(x)^2}}dx=\int\frac{dv}{\sqrt{1+v^2}}=\sinh^{-1}v(x)+C$$ which is what WolframAlpha has done.