Problem

Find the solution of the differential equation th…

01:57

Problem 15 Easy Difficulty

Find the solution of the differential equation that satisfies the given initial condition.
$ x \ln x = y(1 + \sqrt {3 + y^2}) y', y(1) = 1 $

Name: find the solution of the differential equation that satisfies the given initial condition x ln x y1
Uploaded: 2020-03-29
Duration: 1 min 44 s
Channel: Numerade
Description: Find the solution of the differential equation that satisfies the given initial condition. $ x \ln x = y(1 + \sqrt {3 + y^2}) y', y(1) = 1 $

Answer

$$\frac{1}{2} x^{2} \ln x-\frac{1}{4} x^{2}+\frac{41}{12}=\frac{1}{2} y^{2}+\frac{1}{3}\left(3+y^{2}\right)^{3 / 2}$$

Topics

Differential Equations

Calculus: Early Transcendentals

Chapter 9

Differential Equations

Section 3

Separable Equations

Discussion

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Video Transcript

this question asked us to find the solution of the differential equation that satisfies the given initial condition. We know that what we're gonna want to do is put all the left him terms on the left hand side, involving acts and then all the terms involving why on the right hand side, we can split up. Why do you why plus y squared of three y squared And then we know that we have the white, the end. Therefore, we can integrate each of thes three parts separately to give us X squared over to cause X integrates to export over too. And then we have plus C cause we're integrating. Now we know that why have one is one has given the problem. Therefore, we know we can plug in. Why have one is one to give us a sea of negative 41 divided by 12. Therefore, our final solution is the exact same equation we figured out in the previous slide. But instead of pussy, we're gonna be playing in what C is, and it's negative

Amrita B.

Topics

Differential Equations

Calculus: Early Transcendentals

Chapter 9

Differential Equations

Section 3

Separable Equations

Problem