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In Exercises $17-24,$ find the particular solution of the first- order linear differential equationfor $x>0$ that satisfies the initial condition.

$$\begin{array}{ll}{\text { Differential Equation }} & {\text { Initial Condition }} \\ {x d y=(x+y+2) d x} & {y(1)=10}\end{array}$$

$y=x \ln (x)+12 x-2$

Calculus 2 / BC

Chapter 6

Differential Equations

Section 4

First-Order Linear Differential Equations

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Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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In Exercises $17-24,$ find…

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Finding a Particular Solut…

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we have the differential equation X, Do you know why? Equal to X plus why plus two t X and the initial condition. Why at one equal to 10. So first we can rewrite this differential equation Uh, ex d y E v X minus y equal to X plus two. And now So that and now you send the fact that X is greater than zero. We can do by my ex and get the differential equation the wife the ex minus one over x times Why equal to one plus two over X And in this equation, the function p effects is going to be ableto minus one over X and one plus two over X is the function Q over X. So we start computing the integrating factor that is given by the function you off X. It will tow E to the integral off P effects that in this case is minus one over x t X. And that's we know that this integral over here is equal to minus the logarithms off the absolute value effects. So this is going to be ableto eh to the minus absolute valley off the luxury from effects. No, by the same assumption that excess positive is not necessarily to put the absolute body here. Now. Scenes minus the lower rhythm off X is equal to the logarithms off X to the minus one. We get finally that our integrating factor iss X to the minus one or one of our techs. So having this we have the solution is given by the function. Why effects equal to what? Over the integrating factor times the integral off cue effects. And remember that cure fax is one plus two over X times integrating factor the X plus and in the great, um, constancy. So one over one over X is just He's just x times they integral off one over X plus two over X where d x plus c. We has made the problem between one and whatever x and two over X and waterworks. And we know how to integrate each of those terms. So the integral off one over X is the luxury thumb, not real grief off X plus Theis integral here. And this is minus two over X plus c. So here we have that these terms is you're seeing to grow. If one over x d x And this term here is integral off too. Over X were t x. Okay, so that will make the problem with X. And we get X to logarithms off X minus two plus C time six. And this one is the general solution. So don't we use our in the initial condition that why at one has to be able to 10? So we evaluate our our general solution at one, and we will get one times the love Aretha off one minus two plus c times. What now? Whatever. We know that the local rhythm off one is equal to zero. So we get that. Why at one has to be equal to C minus two and this has to be able to 10. So the value off see is 12. So the function Why off x two x logarithms off X uh, plus 12 X minus two. It's a solution off the initial body.

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