We are tasked with solving the following system of differential equations:

$\frac{dx}{dt} = x + t \quad \text{and} \quad \frac{dy}{dt} = x + y$ with the initial conditions: $x(0) = 1, \quad y(0) = 2.$

Step 1: Solve the first equation for $x(t)$

The equation for $x(t)$ is:

$\frac{dx}{dt} = x + t.$

This is a first-order linear differential equation. To solve it:

1. Rewrite the equation: $\frac{dx}{dt} - x = t.$

2. The integrating factor (IF) is: $\mu(t) = e^{\int -1 \, dt} = e^{-t}.$

3. Multiply through by the integrating factor: $e^{-t} \frac{dx}{dt} - e^{-t}x = t e^{-t}.$

The left-hand side is the derivative of $x(t) e^{-t}$: $\frac{d}{dt} \left( x e^{-t} \right) = t e^{-t}.$

4. Integrate both sides with respect to $t$: $x e^{-t} = \int t e^{-t} \, dt.$

To evaluate $\int t e^{-t} \, dt$, we use integration by parts:

• Let $u = t$ and $dv = e^{-t} dt$, so $du = dt$ and $v = -e^{-t}$.

$\int t e^{-t} \, dt = -t e^{-t} - \int -e^{-t} \, dt = -t e^{-t} + e^{-t} + C.$

Thus: $x e^{-t} = -t e^{-t} + e^{-t} + C.$

Multiply through by $e^t$ to isolate $x(t)$: $x(t) = -t + 1 + C e^t.$

Apply the initial condition $x(0) = 1$ to find $C$: $1 = -0 + 1 + C e^0 \implies C = 0.$

Therefore: $x(t) = -t + 1.$

Step 2: Solve for $y(t)$

Now that $x(t) = -t + 1$, substitute it into the second equation:

$\frac{dy}{dt} = x + y \implies \frac{dy}{dt} = (-t + 1) + y.$

Rearrange: $\frac{dy}{dt} - y = -t + 1.$

This is another first-order linear differential equation. The integrating factor is: $\mu(t) = e^{\int -1 \, dt} = e^{-t}.$

Multiply through by the integrating factor: $e^{-t} \frac{dy}{dt} - e^{-t} y = (-t + 1) e^{-t}.$

The left-hand side is the derivative of $y e^{-t}$: $\frac{d}{dt} \left( y e^{-t} \right) = (-t + 1) e^{-t}.$

Integrate both sides with respect to $t$: $y e^{-t} = \int (-t + 1) e^{-t} \, dt.$

Split the integral: $\int (-t + 1) e^{-t} \, dt = -\int t e^{-t} \, dt + \int e^{-t} \, dt.$

We already know from earlier that: $\int t e^{-t} \, dt = -t e^{-t} + e^{-t} + C.$

Thus: $\int (-t + 1) e^{-t} \, dt = -\left( -t e^{-t} + e^{-t} \right) + \int e^{-t} \, dt.$

Simplify: $\int (-t + 1) e^{-t} \, dt = t e^{-t} - e^{-t} - e^{-t} + C.$

Combine terms: $\int (-t + 1) e^{-t} \, dt = t e^{-t} - 2e^{-t} + C.$

Thus: $y e^{-t} = t e^{-t} - 2e^{-t} + C.$

Multiply through by $e^t$ to isolate $y(t)$: $y(t) = t - 2 + C e^t.$

Apply the initial condition $y(0) = 2$ to find $C$: $2 = 0 - 2 + C e^0 \implies C = 4.$

Therefore: $y(t) = t - 2 + 4e^t.$

Final Solution:

The solution to the system of differential equations is: $x(t) = -t + 1, \quad y(t) = t - 2 + 4e^t.$

Do you need further clarifications or additional steps explained?

Here are 5 related questions to expand on this problem:

1. How do you solve systems of differential equations with constant coefficients?
2. What is the method of integrating factors for first-order linear ODEs?
3. Can you solve the same system using a matrix approach?
4. What is the role of initial conditions in solving differential equations?
5. How would the solution change if the equations were nonlinear?

Tip: Always check your solutions by substituting them back into the original equations. This ensures correctness.