To solve this system of initial value problems (IVPs), we have the following equations:
(a) dx₁/dt = -2x₁x₂
dx₂/dt = -3x₂
Let's solve these equations one by one:
1. Start with the second equation:
dx₂/dt = -3x₂
This is a simple first-order linear ordinary differential equation (ODE). The solution to this equation is of the form:
x₂(t) = C * e^(-3t)
where C is a constant that depends on the initial condition for x₂.
2. Now, let's consider the first equation:
dx₁/dt = -2x₁x₂
To solve this, we'll use the fact that we already know x₂(t) from the previous step:
dx₁/dt = -2x₁ * (C * e^(-3t))
This is a separable ODE, so we can rearrange it and integrate:
(1/x₁) dx₁ = -2C * e^(-3t) dt
Integrate both sides:
ln(|x₁|) = 2C/3 * e^(-3t) + K₁
Where K₁ is the constant of integration.
Now, take the exponential of both sides:
|x₁| = e^(2C/3 * e^(-3t) + K₁)
We can write this as:
x₁(t) = A * e^(2C/3 * e^(-3t))
where A is a constant that depends on the initial condition for x₁.
So, the solution to the system of IVPs is:
x₁(t) = A * e^(2C/3 * e^(-3t))
x₂(t) = C * e^(-3t)
Please note that the constants A and C need to be determined based on the initial conditions given for x₁ and x₂.