INSTRUCTOR: Find the general solution to the following differential equations-- part a, the second derivative of a y minus 2 times the first derivative of y plus 10 times y equals 0. With each of these, we're going to consider the characteristic equation. So for this example, part a, the characteristic equation is lambda squared minus 2 lambda plus 10 equals 0.

And we need to solve this with the quadratic formula or with completing the square. But in either case, we can see that lambda will be equal to 1 plus or minus 3i. And because this is a complex solution, this means that the general solution is in this form, y of x equals e to the x, open parentheses, c sub 1 multiplied by cosine open parentheses 3x close parentheses, plus c subscript of 2 times sine, open parentheses 3x close parentheses, close parentheses. And there's our general solution to this because it has complex solutions to the characteristic equation.

Now for part b, we have the equation the second derivative of y plus 14 times the first derivative of y plus 49 times y equals 0. That characteristic equation is lambda squared plus 14 lambda plus 49 equals 0. Now this factors as open parentheses, lambda plus 7 closed parentheses squared equals 0, which means we have a repeated root of lambda equal to negative 7.

Because it's a repeated root, we have a particular form we can use. And that tells our general solution is of this form, y of x equals c subscript of 1 times e to the negative 7x end exponent, plus c sub 2 multiplied by x multiplied by e to the negative 7x end exponent. Again that is because we have a repeated roots of the characteristic equation.