Nonlinear System of Differential Equations

In terms of $p$ and $z$ , the differential equations become $\begin{aligned} \dot{p} & = \; \tfrac23p^2z + \tfrac{7}{10z} \\ \dot{z} & = \; \tfrac13pz^2 + \tfrac{3}{10p} \end{aligned}$ and hence $\tfrac{d}{dt}(pz) \,=\, (pz)^2 + 1$ , which implies that $\tan^{-1}(pz) = t+c$ , for some $c$ , and hence $pz = \tan(t+\tfrac14\pi)$ . But then $\begin{aligned} \dot{p} & = \; \left[\tfrac23\tan(t+\tfrac14\pi) + \tfrac{7}{10}\cot(t + \tfrac14\pi)\right]p \\ \ln p & = \; \tfrac23\ln\big(\sec(t + \tfrac14\pi)\big) + \tfrac{7}{10}\ln\big(\sin(t + \tfrac14\pi)\big) + d \\ p & = \; U \frac{\sin^{\frac{7}{10}}(t + \tfrac14\pi)}{\cos^{\frac23}(t + \tfrac14\pi)} \end{aligned}$ for constants $d,U$ . The initial conditions give us that $U = 2^{\frac{1}{60}}$ . Thus $100ABCDE = 100 \times \tfrac{1}{60} \times 1 \times \tfrac14 \times \tfrac{7}{10} \times \tfrac23 \; =\; \tfrac{7}{36}$ making the answer $\boxed{43}$ .

$\frac{dx}{dt}= \frac{x^2yz}{3} + \frac{0.2}{yz} \ \ \ (1)$ $\frac{dy}{dt} = \frac{xy^2z}{3} + \frac{0.5}{xz} \ \ \ (2)$ $\frac{dz}{dt} = \frac{xyz^2}{3} + \frac{0.3}{xy} \ \ \ (3)$

Multiplying (1) by $yz$ , (2) by $xz$ and (3) by $xy$ and adding leads to:

$\dot{x}yz + x\dot{y}z + xy\dot{z} = (xyz)^2 + 1$ $\implies \frac{d}{dt}(xyz) = (xyz)^2 + 1$

Let: $s = xyz$ . It follows that $s(0) = 1$ .

$\implies \frac{ds}{dt} = s^2+ 1$

Separating the variables, integrating, applying initial conditions and solving for $s(t)$ leads to:

$s(t) = xyz = \tan\left(t + \frac{\pi}{4}\right)$

again consider (1): $\frac{dx}{dt}= \frac{x^2yz}{3} + \frac{0.2}{yz} \ \ \ (1)$ $\implies \dot{x} = \left(\frac{s}{3} + \frac{0.2}{s}\right)x$ $\implies \frac{\dot{x}}{x} = \left(\frac{s}{3} + \frac{0.2}{s}\right)$ $\implies \frac{\dot{x}}{x} =\frac{ \tan\left(t + \frac{\pi}{4}\right)}{3} + 0.2 \cot\left(t + \frac{\pi}{4}\right)$ $\implies \frac{dx}{x} = \left(\frac{ \tan\left(t + \frac{\pi}{4}\right)}{3} + 0.2 \cot\left(t + \frac{\pi}{4}\right)\right) \ dt$ $\implies \int_{1}^{x} \frac{da}{a} =\int_{0}^{t} \left(\frac{ \tan\left(a + \frac{\pi}{4}\right)}{3} + 0.2 \cot\left(a + \frac{\pi}{4}\right)\right) \ da$

$a$ is a dummy variable. Similar steps can be performed to solve for $y$ :

$\implies \int_{1}^{y} \frac{da}{a} =\int_{0}^{t} \left(\frac{ \tan\left(a + \frac{\pi}{4}\right)}{3} + 0.5 \cot\left(a + \frac{\pi}{4}\right)\right) \ da$

Intermediate steps and simplifications are left out in this solution. Multiplying the solutions of $x$ and $y$ leads to:

$p(t) = 2^{1/60} \frac{\sin^{7/10}\left(t + \frac{\pi}{4}\right)}{\cos^{2/3}\left(t + \frac{\pi}{4}\right)}$

And from here, the answer follows. It is $\boxed{43}$ .