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(a) Show that every member of the family of functions $ y = $ (In $ x + C)/x $ is a solution of the differential equation $ x^{2} y^{'} + xy = 1. $

(b) Illustrate part (a) by graphing several members of the family of solutions on a common screen.

(c) Find a solution of the differential equation that satisfies the initial condition. $ y(1) = 2. $

(d) Find a solution of the differential equation that satisfies the initial condition . $ y(2) = 2. $

(a) Solutions of the form $y=\frac{\ln x+C}{x}$ do always solve equations of the form $x^{2} y^{\prime}+x y=1$

(b) See Graph

(c) $y=\frac{\ln x+2}{x}$

(d) $y=\frac{\ln \frac{x}{2}+2}{x}$

Differential Equations

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Idaho State University

Boston College

okay to them going to show you the differential equation. And what is differential equation and the solution off differential equation? And uh huh. And the graph off differential equation. So for the first one, we say here is the differential equation. Yes, it's a equation with a differential y pine, something differential. And for the 1st 1st 1, the solution will satisfy the differential equation. So we just need to calculate clogging to calculate. We put why to hear. And if you calculate directly, you can get this equation here. Also, you can multiple pull the X two, left side, left hand side. I mean the X on the denominator to here and you take the direct differential the D remedy partial derivative for the X. And you get X here in the white pine here by the Prada rule and the and this one is the X equal to 10 x. So this one and this one are equivalent. We just need to beautiful by X two here and his thumb for part B. So we need to grab the why and what when we can see in here. The sea is from minus 5 to 5 so they have a different curve. So in the fees is we have some way called the shooting message to solve the differential equation for the shooting method. We knew the help. They have a certain point we need to pass so we can draw the as many men as we can. So it's very depend on the solution off the differential equation so we can explain more. Percy and D here CNT we just need to solve. So when we have the chance when we have the solution, the solution we have the on nowhere on no constant see here is a constant. So in here in this one if we saw if we knew some point, we can if we knew a point here, we can stop to see So it's canned off like like it's a kind off to solve the sea because we knew the point foot in differential equation. We have a loss off constant, something we need to solve because we don't know which point we need to pass. We just need to parking X equal to one into why, and we get C equal to this. Here is the same thing we can see equal to this. Thank you.

University of California - San Diego

Differential Equations