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: \(y'' - 4y = 5c_{2} e^{-3 t}\)