Let y(t) represent your bank account balance, in dollars, after t years. Suppose you start with $10000 in the account. Each year the account earns 6% interest, and you deposit $5000 into the account. This can be modeled with the differential equation: dy 0.06y + 5000 dt y(0) = 10000 Solve this differential equation for y(t) y(t) =

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### Differential Equation for Bank Account Balance with Interest and Deposits

#### Problem Statement:
Let \( y(t) \) represent your bank account balance, in dollars, after \( t \) years. Suppose you start with $10,000 in the account. Each year the account earns 6% interest, and you deposit $5,000 into the account annually.

This can be modeled with the differential equation:
\[ \frac{dy}{dt} = 0.06y + 5000 \]
\[ y(0) = 10000 \]

#### Solution:
Solve this differential equation for \( y(t) \):

\[ y(t) = \]

Explanation:

1. **Set-Up**: We start with an initial condition where \( y(0) = 10000 \). This means that at time \( t = 0 \), the account balance is $10,000.

2. **Interest**: The account earns 6% interest per year, which translates to a term \( 0.06y \) in the differential equation.

3. **Deposits**: You regularly deposit $5,000 into the account each year, hence the term \( +5000 \) is added to the differential equation.

#### Instructions:
1. Apply techniques for solving first-order linear differential equations.
2. Use the given initial condition to find the particular solution.
3. Provide the final solution in the provided form.

This differential equation models how the balance grows over time considering both interest and annual deposits.
Transcribed Image Text:### Differential Equation for Bank Account Balance with Interest and Deposits #### Problem Statement: Let \( y(t) \) represent your bank account balance, in dollars, after \( t \) years. Suppose you start with $10,000 in the account. Each year the account earns 6% interest, and you deposit $5,000 into the account annually. This can be modeled with the differential equation: \[ \frac{dy}{dt} = 0.06y + 5000 \] \[ y(0) = 10000 \] #### Solution: Solve this differential equation for \( y(t) \): \[ y(t) = \] Explanation: 1. **Set-Up**: We start with an initial condition where \( y(0) = 10000 \). This means that at time \( t = 0 \), the account balance is $10,000. 2. **Interest**: The account earns 6% interest per year, which translates to a term \( 0.06y \) in the differential equation. 3. **Deposits**: You regularly deposit $5,000 into the account each year, hence the term \( +5000 \) is added to the differential equation. #### Instructions: 1. Apply techniques for solving first-order linear differential equations. 2. Use the given initial condition to find the particular solution. 3. Provide the final solution in the provided form. This differential equation models how the balance grows over time considering both interest and annual deposits.
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