💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Like

Report

Numerade Educator

Like

Report

Problem 20 Hard Difficulty

Solve the initial-value problem.
$ (x^2 + 1) \frac {dy}{dx} + 3x(y -1) = 0, y(0) = 2 $

Answer

$$
y=1+\frac{1}{\left(x^{2}+1\right)^{3 / 2}}
$$

Discussion

You must be signed in to discuss.

Video Transcript

In order to get this equation into a form where we can see our P a vaccine or to determine the interpreting factor, we must first divide both sides. In other words, all the terms by the coefficient of y prop, which in this context we know is X squared Plus one. Do this and we get d Y over jacks plus three acts. Divide by ax squared plus one times. Why is three acts again? Divide by X squared, plus one. Okay, Now, or e to the integral of P three acts over X squared plus one D axe. Integrate this. Remember each the natural of X is one. We simply get ax squared, plus one to the power off three divided by two. I'm writing. This is a fractions. It's easier to read. Now. We must multiply all part of the differential equation. So all the terms by or integrating factor and then remember, this is the integral gonna be intruding the right hand side in a second. Okay, getting this, we now get the fact that we can integrate the right hand side. As I said, Okay, we know we can pull out a 1/2 right This is the integral of the squirt of you. Do you know what? Let's say that you is equivalent to X squared plus one. Do you substitution? Right now d'you is two acts DX 1/2 do you is equivalent to x d x. This is U substitution right here, which gives us 1/3 you to the three over two plus c back substitute in What? The value of us. Okay, cool. We have that. Now we know we have. Why is one remember? We want this in terms of why plus c divided by ax squared plus one. The three over to remember original initial value was wives zeros to simply plug in acts. And why, in order to solve for C, the sea is pretty straightforward. Simply one final solution. Plug in our seed into the equation We found out in a previous side and we have our solution