4.5.2

Professor: Example two, "find the general solution for the differential equation xy prime plus 3y equals 4x squared minus 3x, assuming that x is greater than 0." To solve the first order linear differential equation, we are going to follow this problem solving strategy from the text. The first step of which, is put the equation to standard form identifying p and q. 

Let's go ahead and put that right here. This function is almost in standard form. We would need to divide by x everywhere, so that we would have y prime plus 3 over xy times y, and then dividing by x on the right hand side, we'd have 4x minus 3. 

Well, that is in standard form and my p of x, which is actually the more important of the two for this process, is 3 over x. And I'll go ahead and label q of x as 4x minus 3. All right, there's step one. 

Step two, calculate what is going to be called an integrating factor, which is equal to mu of x equals e to the integral, antiderivative of p of x. Let's go ahead and put that in here. So mu of x is e to the antiderivative of 3 over xdx. Well, the antiderivative 3 over x is natural log. 

So this is going to be equal to e to the natural log, or let's write it this way, 3 natural log. That's the value of x. Now, we assumed x is greater than 0, so let's just make that natural log x, which means this is going to be equal to x cubed. That is my integration factor, is x to the third power. 

Our third step is multiply both sides of the differential equation by that mu of x So let's take our y prime, space this out just a little bit. y prime plus 3 over x times y equals 4x minus 3. Multiply both sides by mu of x, so that's going to be the x cubed. 

And that will be x cubed times y prime plus 3x squared times y equals 4x to the 4th. That's 3x cubed. Now, the left hand side is going to be in this form, OK? But this is actually based off the product rule, so we're actually going to have the derivative of that. 

So I'm going to go and write that. This left hand side is the derivative of x cubed times y. And you might look at your product rule and verify if that actually makes sense, but it should. 

Now, when we take the antiderivative of both sides here-- this will be step four. We're going to go and make that with respect to x. The left hand side, by the fundamental theorem of calculus, will just be x cubed times y, and the right hand side will be 4/5x to the 5th, and then minus 3x to the 4th so we have over 4 plus c. 

Last thing to do to find our y, because that's what we're trying to find, is divide by x cubed. So do that. So we have a closed form of y equals 4/5x squared minus 3/4x plus c over x cubed. Now, if we had an initial condition, we could determine what c is. 

[ENDS AT: 5:00]