1000 d x t 25000 1000 4.95 2 2   (a) with given initial conditions: x(0) = 0 and   0 0  t dt dx t (b) (b) Solution for Equation (a) with the specified conditions in Equation (b) is:

The given equation is a differential equation. Let me first rewrite it more clearly:

\[
1000 \frac{dx}{dt} + 25000x = 1000 \cdot 4.95
\]

The initial conditions are:
- \( x(0) = 0 \)
- \( \frac{dx}{dt}(t = 0) = 0 \)

### Step 1: Rewrite the Equation
We can simplify the right-hand side:

\[
1000 \frac{dx}{dt} + 25000x = 4950
\]

Now divide the entire equation by 1000:

\[
\frac{dx}{dt} + 25x = 4.95
\]

### Step 2: Solve the Differential Equation
This is a first-order linear differential equation. To solve it, we'll use an integrating factor method.

The standard form is:

\[
\frac{dx}{dt} + P(t)x = Q(t)
\]

Here, \( P(t) = 25 \) and \( Q(t) = 4.95 \).

The integrating factor \( \mu(t) \) is:

\[
\mu(t) = e^{\int P(t) dt} = e^{25t}
\]

Now, multiply both sides of the differential equation by the integrating factor:

\[
e^{25t} \frac{dx}{dt} + 25e^{25t} x = 4.95 e^{25t}
\]

The left side is the derivative of \( e^{25t} x \), so:

\[
\frac{d}{dt} \left( e^{25t} x \right) = 4.95 e^{25t}
\]

### Step 3: Integrate Both Sides
Now, integrate both sides with respect to \( t \):

\[
e^{25t} x = \int 4.95 e^{25t} dt
\]

The right-hand side integrates as:

\[
\int 4.95 e^{25t} dt = \frac{4.95}{25} e^{25t} = 0.198 e^{25t}
\]

So, we have:

\[
e^{25t} x = 0.198 e^{25t} + C
\]

Where \( C \) is the constant of integration. Divide by \( e^{25t} \) to solve for \( x \):

\[
x(t) = 0.198 + C e^{-25t}
\]

### Step 4: Apply Initial Conditions
We are given that \( x(0) = 0 \). Substituting \( t = 0 \) into the equation:

\[
0 = 0.198 + C e^0
\]

This gives:

\[
0 = 0.198 + C
\]

So \( C = -0.198 \).

Thus, the solution to the differential equation is:

\[
x(t) = 0.198 - 0.198 e^{-25t}
\]

### Final Solution:
The solution for the given differential equation with the specified initial conditions is:

\[
x(t) = 0.198(1 - e^{-25t})
\]

Would you like further clarification on any step?

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**5 Related Questions:**
1. How do integrating factors work in solving linear differential equations?
2. What are other methods to solve first-order differential equations?
3. How do we apply initial conditions to determine constants in solutions?
4. How does the solution behavior change as \( t \to \infty \)?
5. What are real-world applications of differential equations like this one?

**Tip:** Always check the initial conditions after solving a differential equation to ensure your solution is correct.

Learn how to solve a first-order linear differential equation 1000 dx/dt + 25000x = 1000 * 4.95 with the initial conditions x(0) = 0 and dx/dt(0) = 0. Step-by-step explanation using the integrating factor method.

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