Solve the initial value problem.
y prime equals 0.05 y minus 48 , y(0)=500

To solve the initial value problem (IVP):

\[
\frac{dy}{dt} = 0.05y - 48, \quad y(0) = 500
\]

we'll solve the differential equation step by step.

### Step 1: Write the equation in a standard form
The equation can be rewritten as:

\[
\frac{dy}{dt} = 0.05y - 48
\]

This is a linear first-order differential equation.

### Step 2: Find the integrating factor
The standard form of a linear first-order differential equation is:

\[
\frac{dy}{dt} + p(t)y = q(t)
\]

For this equation, \( p(t) = -0.05 \) and \( q(t) = -48 \).

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

\[
\mu(t) = e^{\int p(t) \, dt} = e^{\int -0.05 \, dt} = e^{-0.05t}
\]

### Step 3: Multiply the entire differential equation by the integrating factor
Multiply both sides of the differential equation by \( e^{-0.05t} \):

\[
e^{-0.05t} \frac{dy}{dt} - 0.05e^{-0.05t}y = -48e^{-0.05t}
\]

This can be rewritten as:

\[
\frac{d}{dt} \left( y \cdot e^{-0.05t} \right) = -48e^{-0.05t}
\]

### Step 4: Integrate both sides
Integrate both sides with respect to \( t \):

\[
y \cdot e^{-0.05t} = \int -48e^{-0.05t} \, dt
\]

To integrate the right-hand side, recognize that the integral of \( e^{-0.05t} \) is \( -\frac{1}{0.05} e^{-0.05t} \):

\[
y \cdot e^{-0.05t} = \frac{48}{0.05} e^{-0.05t} + C
\]

Simplify:

\[
y \cdot e^{-0.05t} = 960 e^{-0.05t} + C
\]

### Step 5: Solve for \( y(t) \)
Multiply both sides by \( e^{0.05t} \) to solve for \( y(t) \):

\[
y(t) = 960 + Ce^{0.05t}
\]

### Step 6: Apply the initial condition
Use the initial condition \( y(0) = 500 \) to find \( C \):

\[
500 = 960 + C \cdot e^{0}
\]

\[
C = 500 - 960 = -460
\]

### Step 7: Write the final solution
Substitute \( C = -460 \) back into the general solution:

\[
y(t) = 960 - 460e^{0.05t}
\]

This is the solution to the initial value problem.

Would you like further details or have any questions? Here are 5 related questions for practice:

1. Solve the initial value problem \( \frac{dy}{dt} = 0.1y - 30 \), \( y(0) = 200 \).
2. Find the general solution to \( \frac{dy}{dt} = 0.03y + 50 \).
3. What is the long-term behavior of the solution as \( t \) approaches infinity?
4. Solve \( \frac{dy}{dt} = 0.05y \) with the initial condition \( y(0) = 100 \).
5. Determine the equilibrium solution of \( \frac{dy}{dt} = 0.05y - 48 \).

**Tip:** Always check the consistency of your solution by substituting it back into the original differential equation.

Solve the initial value problem. y' = 0.05y - 48, y(0) = 500

This page provides a step-by-step solution to the initial value problem y' = 0.05y - 48 with the initial condition y(0) = 500. Learn about solving linear first-order differential equations using the integrating factor method, with a focus on understanding key concepts like integrating factors and applying initial conditions. Ideal for undergraduate students studying Calculus.

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