A frigate originally located in the plane moves at constant speed a in the positive direction along the vertical axis. At the same time, a galleon at the point of coordinates $(c, 0)$ sets off in pursuit (for a collision) at constant speed b. Show, using the fact that the galleon is always moving in the direction of the frigate, that the position $(x, y)$ of the galleon satisfies the equation.
$$x \dfrac {d^2 y}{dx^2} = \dfrac ab \sqrt {1+\left(\dfrac {dy}{dx}\right)^2}$$
Solve this equation, considering cases $a > b$ and $a < b$ separately
enter image description here This is a picture of a part of homework I have been stuck for a full day and I don't now how to proceed.
Note:
$\dfrac {d^2 y}{dx^2}$ = second derivative of $y$; $\dfrac {dy}{dx}$ = first derivative of y
The equation being $$x y'' = \dfrac ab \sqrt {1+\left(y'\right)^2}$$ reduce the order $p=y'$ to get $$x p'=\dfrac ab \sqrt {1+p^2}$$ which is separable. So $$p=\sinh \left(\frac{a \log (x)}{b}+ c_1\right)=\sinh(c_1) \cosh \left(\frac{a \log (x)}{b}\right)+\cosh (c_1) \sinh \left(\frac{a \log (x)}{b}\right)$$ So, now, you need to compute $$I=\int \cosh \left(\frac{a \log (x)}{b}\right)\,dx\qquad \text{and}\qquad J=\int \sinh \left(\frac{a \log (x)}{b}\right)\,dx$$ Let $$\frac{a \log (x)}{b}=t \implies x=e^{\frac{b t}{a}}\implies dx=\frac{b }{a}e^{\frac{b t}{a}}\,dt$$ $$I=\frac{b }{a}\int e^{\frac{b t}{a}}\cosh(t)\,dt\qquad \text{and}\qquad J=\frac{b }{a}\int e^{\frac{b t}{a}}\sinh(t)\,dt$$ Compute $(I+J)$ and $(I-J)$ to face simple integrals using $\cosh(t)\pm\sinh(t)=e^{\pm t}$