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Numerade Educator

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Problem 3 Easy Difficulty

Solve the differential equation.
$ xyy' = x^2 + 1$

Answer

$y=\pm \sqrt{x^{2}+2 \ln |x|+C}$ where $C=2 K$

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Video Transcript

So the differential equation that we're going to solve here is X times why times wine prime equals X squared. Plus what? This is a first order differential equation. The derivatives only appear up to first order in any of the variables in this equation. There are no second order or third order derivatives here, so it's a first order equation and it is separable, meaning that the X factors in terms will be able to be separated over to one side of the equation and why factors in terms will remain on the other side of the equation. This will allow us to apply separation of variables in order to solve the equation. Now to get started. Let's rewrite this equation in live news notation. So we get X times Why times D y e X equals X squared plus one. Lively's notation is very convenient with separation of variables, so we're going to stick with that for this exercise. Now, to begin the process of separating the variables in this equation where we're going to do is divide both sides of the equation by ex That gives us why times d y d X equals x squared plus one over X, with the end result being that all the X factors in terms are over on the right side of the equation and the wife actors in terms are on the left side of the equation now, the next step here is to rewrite this equation in differential form. What we're going to do is to multiply both sides of the equation by this differential d x, so to speak, giving us why times d y equals X squared plus one over x d X. It's not exactly rigorous, but it is a convenient framework for applying separation of variables. And it does give us the correct answer in the end, which is the most important thing here. The only thing to keep in mind is that this derivative D. Y. D X is not actually a quotient. No, The next step here is to integrate both sides of the equation. So the integral of why d y equals the integral of X squared plus one over X d x. And it's critical here that we were able to isolate these variables on the one side of the equation over the other, because that's what allows us to do these in a girl's on each side. Now the integral of y D Y is very straightforward. It's just why squared over two and we're going to have to add a constant, but we'll do that on the other side of the equation. The integral of X squared plus one over X is also not so bad. It's just going to turn out, ah, to distribute over this numerator. So we get X squared over X plus one over X d x, which is equal to the integral of X plus one over X d x and the integral distributes over the sum. So we get the integral of exit e X plus the integral of one over X, the ex the integral of X t X is just X squared over two. The integral of one over X is the natural log rhythm of the absolute value of X, and now we're finally going to add a constant. So now that we've evaluated this integral, what we can do is bring it over to the right hand side of this equation and get X squared over two plus Ln of the absolute value of X plus some constant C. Now we can multiply both sides of the equation by two and get y square equals X squared, plus two times Ellen of the absolute value of X plus two times seat now two times some constant is really just some other constant, so it's not really necessary to go to the trouble of renaming this thing. We're just gonna call it see again now. The last step here is to take square root of both sides, so that gives us why is equal to plus or minus the square root of X squared, plus two times L and of absolute value of X plus C, and that is the solution.