The system of differential equations provided is:

$\frac{dx}{dt} = x \quad \text{with} \quad x(0) = 1$ $\frac{dy}{dt} = y \quad \text{with} \quad y(0) = -4$

Both equations are separable and can be solved independently.

1. Solve $\frac{dx}{dt} = x$:

We separate variables and integrate:

$\frac{1}{x} dx = dt$ $\int \frac{1}{x} dx = \int dt$ $\ln|x| = t + C_1$ Exponentiate both sides: $x = e^{t + C_1} = e^{C_1} e^t$ Let $e^{C_1} = C$, then: $x(t) = Ce^t$ Using the initial condition $x(0) = 1$: $1 = Ce^0 \quad \Rightarrow \quad C = 1$ Thus: $x(t) = e^t$

2. Solve $\frac{dy}{dt} = y$:

We follow the same process:

$\frac{1}{y} dy = dt$ $\int \frac{1}{y} dy = \int dt$ $\ln|y| = t + C_2$ Exponentiate both sides: $y = e^{t + C_2} = e^{C_2} e^t$ Let $e^{C_2} = D$, then: $y(t) = De^t$ Using the initial condition $y(0) = -4$: $-4 = De^0 \quad \Rightarrow \quad D = -4$ Thus: $y(t) = -4e^t$

Final Solutions:

$x(t) = e^t$ $y(t) = -4e^t$

Would you like further details on the solving process or have any questions?

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Here are 5 related questions for deeper understanding:

1. What is a separable differential equation?
2. Can you solve $\frac{dz}{dt} = z$ with $z(0) = 5$?
3. How do we interpret the solutions $x(t) = e^t$ and $y(t) = -4e^t$ geometrically?
4. What happens if the initial conditions are different for $x(0)$ and $y(0)$?
5. How would the solutions change if the differential equations were nonlinear?

Tip: Always check your initial conditions after solving differential equations to ensure the constants are correctly determined.