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The differential equation provided is of the form:

\[ y(t) = A e^{2t} + B e^{-3t} \]

where \( A \) and \( B \) are constants.

We are given the initial conditions:
\[ y(0) = 7 \]
\[ y'(0) = 2 \]

First, we need to calculate \( y'(t) \) from \( y(t) \):
\[ y'(t) = 2A e^{2t} - 3B e^{-3t} \]

Using the initial condition \( y(0) = 7 \):
\[ y(0) = A e^{0} + B e^{0} = A + B = 7 \]

Using the initial condition \( y'(0) = 2 \):
\[ y'(0) = 2A e^{0} - 3B e^{0} = 2A - 3B = 2 \]

Now we have a system of linear equations:
1. \( A + B = 7 \)
2. \( 2A - 3B = 2 \)

Let's solve this system to find the values of \( A \) and \( B \).The constants \( A \) and \( B \) are found to be:
\[ A = \frac{23}{5} \]
\[ B = \frac{12}{5} \]

Substituting these values back into the equation for \( y(t) \), the solution to the differential equation is:
\[ y(t) = \frac{23}{5} e^{2t} + \frac{12}{5} e^{-3t} \]

Would you like to know more details about any step or have any other questions?

Here are some follow-up questions you might consider:
1. How would the solution change if the initial conditions were different?
2. What happens to \( y(t) \) as \( t \) approaches infinity?
3. Can you explain why the coefficients of \( e^{2t} \) and \( e^{-3t} \) in the solution have different signs?
4. What would be the solution if only the initial condition \( y'(0) \) was given?
5. What are the implications of a solution like this in real-world applications?

Tip: Understanding the exponential terms in the solution can provide insight into the behavior of the system, such as whether the system is stable or unstable.

The solution of a certain differential equation is of the form y(t) = A * e^(2t) + B * e^(-3t), where A and B are constants. The solution has initial conditions y(0) = 7 and y'(0) = 2. Find the solution.

Learn how to solve a differential equation of the form y(t) = A * e^(2t) + B * e^(-3t) using initial conditions y(0) = 7 and y'(0) = 2. This problem involves exponential functions and solving linear systems.

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