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.

dx/dt = x, dy/dt = y with x(0) = 1, y(0) = -4

Explore how to solve the system of differential equations dx/dt = x and dy/dt = y with initial conditions x(0) = 1 and y(0) = -4. Learn the step-by-step process using separation of variables and the exponential growth formula. Suitable for Grades 10-12.

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