“Seek-Lock-Strike!” Simplified

There is an easier way to derive the governing equation ((5), “Seek-Lock-Strike!“) for the missile.

Solving

$\frac{dy}{dx}(a+v_at-x) = b-y$

for $t,$ we have

$t = \frac{(x-a)\frac{dy}{dx}-y+b}{v_a\frac{dy}{dx}}.\quad\quad\quad(1)$

From

$v_m t = \int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx,$

we also have

$t = \frac{\int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx}{v_m}.\quad\quad\quad(2)$

Equate (1) and (2) gives

$\frac{\int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx}{v_m}-\frac{(x-a)\frac{dy}{dx}-y+b}{v_a\frac{dy}{dx}}=0.\quad\quad\quad(3)$

Differentiate (3) with repect to $x,$ we obtain

$(bv_m-v_my)\frac{d^2y}{dx^2} + v_a(\frac{dy}{dx})^2\sqrt{1+(\frac{dy}{dx})^2} = 0.\quad\quad\quad(4)$

(see Fig. 1)

Fig. 1

Let $r=\frac{v_a}{v_m}, p=\frac{dy}{dx} \implies \frac{d^2y}{dx^2} = p\frac{dp}{dy},$ (4) bcomes

$(b-y)p\frac{dp}{dy} + r p^2 \sqrt{1+p^2} = 0.$

Since $p = \frac{dy}{dx} \ne 0$, dividing $p$ through yields

$(b-y)\frac{dp}{dy} + rp\sqrt{1+p^2}=0,$

the governing equation for the missile.