Problem

Solve the given differential equation. $y^{\prim…

03:06

Problem 13 Easy Difficulty

Solve the given differential equation.
$$y^{\prime}-x^{-1} y=2 x^{2} \ln x$$

Answer

$x^{3} \ln (x)-\frac{x^{3}}{2}+c_{1} x$

Related Courses

Calculus 2 / BC

Differential Equations and Linear Algebra

Chapter 1

First-Order Differential Equations

Section 6

First-Order Linear Differential Equations

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Video Transcript

Hello. Let's get to solve its problems, shall we? So only to do is do you find integrating factor from this position. So integrating factor is equal to e to the integral of p of X, which is the function in front of why. So it's going to be a negative. Eat the one over X negative e to the one over the integral of one cracks, which sounds like a mouthful. But it's literally just this. So we just take the integral of each of the one of Rex. Sorry, Integral of one of Rex, which is gonna be, um, Alan of X. Then finally, the minus. And already there's we're gonna pull the minus sign up to the top right here. So you know, it would e to the Ellen of X to the minus one, which give you to eat of the island of something you end up with just what's in front of the brackets of the L Innovex. Sorry. In front, back is Ellen, which you end up with one of her ex. Okay, so now we need to multiply both sides by integrating factor way. So we end up with one of her acts why prime minus one over X Y is equal to one over x times two x squared which, if you multiply x squared by one of the extras and with two X and the l a X So now we gotta integrate both sides left inside It just turns into white times an integrating factor because if you take the integral of a derogative your brackets girl one. So this one's different of is right here. So you take it's integral narrative. You in a packet. Score one Now, is it in a row off this part? To do this, we need integration by parts. So we're gonna let this part B You let this part B D v formulas UV minus in agro signed eu times of V Okay, so we end up with you, which is Ellen of X Times DV, which is integral of two extra typical two x squared minus integral sign derivative. Ellen X is one over x times V, which is gonna be X squared. These cancel and give us just x snow edited the general just x well out here in a bit X squared Ellen up X said integral of X is going to be X squared over two and then being a put a constant of plus c Okay, So now always to do is multiply both sides by X to get rid of that one of her ex. So we get why is equal to X cubed Ellen X minus X cute over to plus C

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