Question

Find a differential equation whose general solution is \(y=c_{1} e^{2 t}+c_{2} e^{-3 t}\)

Short answer

In this exercise, we found that the given general solution \(y=c_{1} e^{2 t}+c_{2} e^{-3 t}\) satisfies a second-order differential equation \(y'' - 4y = 5c_{2} e^{-3 t}\). This means that the general solution can be used to solve this differential equation for any values of the constants \(c_1\) and \(c_2\).

Step-by-step solution

Determine the order of the differential equation

Since the given solution has two linearly independent functions (\(e^{2 t}\) and \(e^{-3 t}\)), the order of the differential equation must be at least 2. That means we need to find a relationship between \(y\), \(y'\), and \(y''\).

Compute the first derivative of the function y

To compute the first derivative \(y'\), we will differentiate each term of the general solution \(y=c_{1} e^{2 t}+c_{2} e^{-3 t}\) with respect to \(t\). \(y' = \frac{d}{dt}(c_{1} e^{2 t})+\frac{d}{dt}(c_{2} e^{-3 t})\) Using the chain rule, we get: \(y' = 2c_{1} e^{2 t}-3c_{2} e^{-3 t}\)

Compute the second derivative of the function y

Doing the same process, we will compute the second derivative \(y''\) by differentiating the first derivative \(y'\) with respect to \(t\). \(y'' = \frac{d}{dt}(2c_{1} e^{2 t})+\frac{d}{dt}(-3c_{2} e^{-3 t})\) Using the chain rule, we get: \(y'' = 4c_{1} e^{2 t}+9c_{2} e^{-3 t}\)

Find a relationship between y, y', and y''

Now, we need to find a relationship between \(y\), \(y'\), and \(y''\). We can try to eliminate either \(c_1\) or \(c_2\) by algebraic manipulations and see if such an elimination gives us a reasonable differential equation. We will try to eliminate \(c_1\) by multiplying \(y\) by a constant and then subtracting it from the second derivative \(y''\). Let's multiply \(y\) by 4: \(4y = 4c_{1} e^{2 t}+4c_{2} e^{-3 t}\) Now, by subtracting 4 times \(y\) from \(y''\): \(y''-4y = 4c_{1} e^{2 t}+9c_{2} e^{-3 t} - (4c_{1} e^{2 t}+4c_{2} e^{-3 t})\) After simplification, we get: \(y''-4y = 5c_{2} e^{-3 t}\) While we didn't succeed in eliminating both constants, we were able to eliminate \(c_{1}\) from the equation.

Conclusion

The given general solution \(y=c_{1} e^{2 t}+c_{2} e^{-3 t}\) can satisfy the following second-order differential equation: \(y'' - 4y = 5c_{2} e^{-3 t}\)

Key concepts

General Solution

The concept of a general solution in differential equations is fundamental. It represents a family of solutions that encompass all possible specific solutions for a differential equation. In the context of the exercise, the general solution is given as \(y = c_1 e^{2t} + c_2 e^{-3t}\). Here, \(c_1\) and \(c_2\) are arbitrary constants that allow the solution to be adjusted to fit specific initial conditions.

When solving differential equations, finding the general solution is often the ultimate goal, as it provides the most complete picture of the system's behavior. It's important to note that general solutions exist for different types of differential equations, including ordinary and partial differential equations.

Understanding how to manipulate and work with a general solution is crucial for accurately describing dynamical systems and predicting future behaviors.

Linearly Independent Functions

Linearly independent functions play a key role in forming the general solutions of differential equations. In our example, the functions \(e^{2t}\) and \(e^{-3t}\) are linearly independent. This means they do not "overlap" or copy each other's behavior over any interval, essentially providing unique contributions to the solution.

The concept of linear independence is vital because it ensures that the solution space is fully spanned.

• Two functions, \(f(t)\) and \(g(t)\), are said to be linearly independent if no constant \(k\) exists such that \(f(t) = k g(t)\) for all \(t\).
• In other words, you can't express one in terms of the other through simple scaling, not involving zero.

This characteristic guarantees a second-order differential equation has a full solution set which is crucial when interpreting complex systems or solving real-world problems. If the functions were linearly dependent, they wouldn't provide sufficient information to solve the equation properly.

Chain Rule

The Chain Rule is a fundamental technique in calculus used to differentiate composite functions. It is utilized extensively in solving differential equations, especially when you need to find first and second derivatives, as seen in this exercise.

For a given function \(h(t)\) that can be expressed as \(h(t) = f(g(t))\), the chain rule provides a way to find the derivative as: \(\frac{dh}{dt} = f'(g(t)) \cdot g'(t) \)

In the context of the provided general solution, the chain rule was applied to differentiate the exponential functions.

• For the term \(c_1 e^{2t}\), differentiate according to \(2t\) which gives \(2c_1 e^{2t}\).
• Similarly, for \(c_2 e^{-3t}\), it yields \(-3c_2 e^{-3t}\).

Mastery of the chain rule enables solving complex differential equations by accurately breaking down and understanding how each part of the data set changes with respect to time or another variable.

Second-Order Differential Equation

Second-order differential equations involve derivatives up to the second degree. The exercise demonstrates the derivation of such an equation from a general solution with a specific structure of linearly independent functions.

When converting a general solution like \(y = c_1 e^{2t} + c_2 e^{-3t}\) into a corresponding differential equation, you typically seek relationships between \(y\), \(y'\), and \(y''\) to determine the order. In this case, because two linearly independent functions are involved, we expect a second-order equation.

• The presence of \(y''\) as the highest derivative directly signals it's a second-order problem.
• This gives them particular characteristics, including oscillatory solutions, damping, and more.

Understanding these equations is crucial since they describe a range of physical phenomena, including spring motion, wave behavior, and electronic circuits. Solving second-order differential equations thus provides insights into predicting such dynamics over time.