INSTRUCTOR: Find the general solution to the following differential equations. In this question, there are two parts, and in this video we will only be solving part b, which is the second derivative of x minus 2 times the first derivative of x plus x equals a fraction. Numerator is e to the t, and the denominator is t.

So let's consider the complementary equation. That is the second derivative of x minus 2 times the first derivative of x plus x equals 0. The characteristic equation of that complementary equation would be lambda squared minus 2 lambda plus 1 equals 0, from which we can see that lambda is equal to 1, and it's a repeated solution. So that the general solution-- general solution will be of the form c1 times e to the t plus c2 times t times e to the t.

Now we're going to set up a system of equations. Again, this is called the method of variation of parameters. From the general solution, I'm going to say that y1 is e to the t, and y2 is t times e to the t. So variation of parameter says we'll set up a system which is u prime-- u prime times y1-- so times e to the t-- plus v prime times y2, which is t e to the t, equals 0.

And the second equation is u prime times y1 prime plus v prime times y2 prime equals r of x, which is our e to the t all over t. So that leaves us with u prime e to the t plus v prime, open parenthesis, e to the t plus t times e to the t, close parenthesis, equals our fraction-- e to the t as the numerator and t as the denominator.

Now, we could use any number of methods, including Cramer's rule. That's a very popular one. But I noticed that two of these terms could actually be eliminated very easily. So if we take the second equation and subtract the first-- take the second equation, subtract the first-- this leaves us with an equation that is v prime times e to the t equals e to the t as the numerator divided by t. Which means that v prime is equal to 1 over t.

And finding the antiderivative, v is equal to natural log absolute value of t, end absolute value. And then, of course, plus c, but that's not going to affect our solution, so I'm not going to worry about that.

Now, from this-- from this, we can take the-- we can actually evaluate that into the first equation-- that's our first equation-- as u prime times e to the t plus v prime, which I'm going to substitute in as 1 over t, times t times e to the t. That equals 0. And we could solve this for u prime. u prime is equal to negative 1, which means that u, finding our antiderivative, will be equal to negative x. Actually, negative t. We changed variables there. Use the correct one.

Now, taking our general solution, and now this y sub p that we can deduce from this, we can write that y-- I need to use the right variable. This is x of t equals c sub 1 times e to the t plus c sub 2 times t times e to the t minus t e to the t plus natural log absolute value of t, end absolute value, times t times e to the t, using our u and our v as a c1 and c2 of sorts for that particular solution.

One thing remains, and that is the we can combine two terms here. We have c sub 2 times t e to the t minus 1t e to the t. We can combine that as a constant and say that x of t is equal to c sub 1 times e to the t plus c sub 2 times t times e to the t plus ln absolute value of t, end absolute value, times t times e to the t.