(Solved): Consider a conflict between two armies of \( x \) and \( y \) soldiers, respectively. During World ...

Consider a conflict between two armies of \( x \) and \( y \) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiors in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \( t \) represents time since the start of the battle, then \( x \) and \( y \) obey the differential equations \[ \frac{d x}{d t}=-a y, \quad \frac{d y}{d t}=-b x_{1} \] where \( a \) and \( b \) are positive constants. Suppose that \( a=0.04 \) and \( b=0.01 \), and that the amies start with \( x(0)=58 \) and \( y(0)=17 \) thousand soldiers. (Use units of thousands of soldiers for both \( x \) and \( y \).) (a) Rewrite the system of equations as an equation for \( y \) as a function of \( x \) : \( \frac{d y}{d x}= \) (b) Solve the differential equation you obtained in (a) to show that the equagtion of the phase trajectory is. \[ 0.04 y^{2}-0.01 x^{2}=C \text {, } \] for some constant \( C \). This equation is called Lanchester's square law. Given the initial conditions \( x(0)=58 \) and \( y(0)=17 \), what is \( C \) ? \[ C= \]