INSTRUCTOR: Find the general solution to the following differential equations. In this question, there are two parts, a and b. And in this video, we will only solve part a, which is the second derivative of y, plus y, equals secant x.

First, let's consider the complementary equation, which would be the second derivative of y, plus y, equals 0. And the characteristic equation of that complementary equation would be lambda squared plus 1 equals 0, which means that lambda equals plus or minus i.

Now this tells us that the general solution is of the form c sub 1, times e-- oh, wrong question. This tells us that the general solution is of the form c sub 1 times cosine x, plus c sub 2 times sine x. Next we will set up a system of equations. And by the way, this is called the method of variation of parameters.

So based on that general solution, we'll set up a system of equations, and it is going to be u prime, times y sub 1, which is actually the cosine of the general equation. So u prime, cosine x, plus v prime, times sine x, equals 0.

And for our second equation, we're going to take the derivative of the y1 and y2 too. We have the cosine and the sine respectively. And it'll be equal to the other side of the equation. So this will be negative u prime, sine x, plus v prime, times cosine x, equals secant x.

Now we can solve this in a number of ways, but I'm going to go ahead and use Cramer's rule, which tells us that we actually consider certain columns of this as a matrix, certain variable sets. Go back and review Cramer's rule if you need to, but I'll go and write what this would be.

Cramer's rule would say that u prime is going to be equal to a fraction, and the numerator is the determinant of a matrix, and the denominator is a determinant of a matrix. The numerator matrix will have the first row of 0 and then sine x. The second row is secant x and then cosine x. The denominator matrix will have a first row of cosine x and then sine x, and a second row of sine x and then cosine x.

Now finding the determinants of these two matrices, in the numerator yields negative tangent x, and in the denominator, we have 1. So the new prime is equal to negative tangent x.

Now that means, finding our intuitive, that u is equal to negative natural log, absolute value of secant x and absolute value. Alternatively, that is equal to the natural log of the absolute value of cosine x and absolute value, plus c for our constant of integration, which won't come up in our solution.

Also, Cramer's rule tells us how to find v as well in the system. By Cramer's rule, v prime is equal to a fraction, numerator and denominator are both determinants of 2 by 2 matrices. The numerator matrix has a first row of cosine x and then 0, and a second row of sine x and then secant x.

The denominator matrix is the first row of cosine x and then sine x, and a second row of sine x and then cosine x. Finding those determinants respectively, the numerator is 1, and the denominator is also 1. So that v prime is equal to 1, which means that v, finding range derivative, is equal to x, plus c. But again that's not going to affect our final solution.

So knowing these things, we're going to sum our general solution, and this other solution with the u and v values in place of c1 and c2 respectively. So y of x equals c sub 1 times cosine x, plus c sub 2, times sine x, plus natural log absolute value, cosine x and absolute value, times cosine x, plus x, times sine x.