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Solve the following ivps. (a) dx₁ dt = −2x₁ x₂ dx₂ dt = 0x₁ − 3x₂

User Ritu Dhoot
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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.
User Rfb
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