INSTRUCTOR: Consider the differential equation second derivative of y plus 5 times the first derivative of y plus 6 times y equals 0. Given that e to the exponent of negative 2x, end exponent, and e to the negative 3x, end exponent, are solutions to this differential equation. Show that 3e to the negative 2x, end exponent, plus 6e to the negative 3x, end exponent, is a solution.

So let's consider this function as y equals 3e to the negative 2x, end exponent, plus 6e to the negative 3x, end exponent. From that, we can see that the first derivative of y is equal to negative 6e to the negative 2x, end exponent, minus 18e to the negative 3x, end exponent. And also, the second derivative of y, equal to 12e to the negative 2x, end exponent, plus 54e to the negative 3x, end exponent.

Now, if we take our second derivative plus 5 times our first derivative, plus 6 times the function itself, y, that should be equal to 0 if it's a solution. So that would be 12e to the negative 2x, end exponent, plus 54e to the negative 3x, end exponent, plus 5 open parentheses, negative 6e to the negative 2x, end exponent, minus 18e to the negative 3x, end exponent, close parentheses, plus 6, open parentheses, 3e to the negative 2x, end exponent, plus 6e to the negative 3x, end exponent, close parentheses.

And if we start equating coefficients here, the e to the negative 2x terms have 12 minus 30 plus 18. That quantity times e to the negative 2x, and 12 minus 30 plus 18 is in fact 0. So we have 0e to the negative 2x. And then the plus open parentheses, 54 minus 90 plus 36, close parentheses, e to the negative 3x, end exponent. And that coefficient is also equal to 0. So this entire expression is in fact 0, which means that this function is a solution to that differential equation.