Answer to Question #156210 in Differential Equations for Rashinda

Question

Solve the simultaneous differential equations dx/dt + ky = 0 and dy/dt + kx = 0 with the initial conditions that x = 0 and y = 1 when t = 0. Hence, show that for all values of t, y^2 - x^2 = m where m is a constant to be determined.

Answer 1

Solution

Differentiating first equation on t and substituting dy/dt from second equation we will get equation for x(t)

d²x/dt²-k²x = 0

Characteristic equation for it is λ²-k²=0. => λ₁=k, λ₂=-k. Therefore solution for x(t) is

x = A*e^k*t+B*e^-k*t

From initial condition x = 0 when t = 0. => A+B = 0 => B = -A => x = A*( e^k*t-e^-k*t) = 2*A*sinh(k*t)

From first equation y = -(1/k)* dx/dt = 2*A*(1/k)* d sinh(k*t) /dt = 2*A*cosh(k*t)

From initial condition y = 1 when t = 0. => 2*A = 1 => A = 1/2

So solution of the system of differential equations is x = sinh(k*t), y = cosh(k*t)

According to properties of hyperbolic functions (https://en.wikipedia.org/wiki/Hyperbolic_functions)

y²-x² = cosh²(k*t) - sinh²(k*t) = 1 => m = 1

Answer

x = sinh(k*t), y = cosh(k*t), m = 1