The Graph of a Member of a Family of Solutions of a Second-order Differential Equation

The graph of member of family of solutions of second-order differential equation dyldx2 {x, Y, %) is given. Match the solution curve with at least one pair of the following Initial weather _ (Select all that apply.)Y(1) = ane,Y(1) = -2 y(-1) = 0,Y"(-1) = y(1) 1,Y'(one) = 2 Y(0) -1,Yt0) = 2 Y(0) ~ane, Y"(0) Y(0) -4, Y"0)

The graph of member of family of solutions of second-order differential equation dyldx2 {10, Y, %) is given. Match the solution curve with at least i pair of the following Initial weather _ (Select all that utilize.) Y(1) = 1,Y(1) = -2 y(-ane) = 0,Y"(-1) = y(1) i,Y'(1) = 2 Y(0) -one,Yt0) = 2 Y(0) ~1, Y"(0) Y(0) -iv, Y"0)

Discover the family unit of solutions of the given differential equation and
the family of orthogonal trajectories. Sketch both families.
$$
\brainstorm{array} { fifty } { \text { a. } ten d x + y d y = 0 } \\ { \text { b. } x d y - two y d x = 0 } \cease{array}
$$

In this video, nosotros're gonna get through the respond to question number 12 from affiliate xvi.three. And so a for united states to observe the family of solutions off this unlike to the question on let's find the family of all doctoral trajectories. First, to observe the solutions, we tin can but, ah separate the variables into X. The X is equal to minus why you lot why and integrate both sides. We demand to add together a constant of integration. We're gonna accept half X squared. Allow's put the wise and on the left inside as well. Plus 1/2 why? Swears is able to come across allow me merely most by both sides past two and blot the gene 2 into C considering information technology'south just are pretty constant. Who got X where y squared plus X squared equal to encounter. And then we're now we demand to find the authorization of trajectories at this. Cheers. Why the 10 for these curves? He's gonna exist acquitted minus X over. Why information technology took me to encounter that straight from the difference equation that were given so that for the All Star y'all know, trajectories, we're gonna have why the X equal to the negative Pacifica with this, which is why of 10. Then we could separate the very was again. You got one of why why equals in schoolhouse of ane of 10 dx, plus a constant of integration? Do you lot have the natural log? Why is equal to the natural log of X? Nosotros could take East to the power of both sides. So information technology'due south essential that both sides, they even have why is equal to a constant a Times X sees your straight lines through the origin and so you tin meet that the solutions we're circles considering they were X squared plus y squared is equal toe. See, So the seconds would various radio radio I on the orthogonal trajectories, straight lines through the origin like second, I'thousand not be whereas to notice these lesions of this defense force situation a split up the very bulls and nosotros're gonna have any. Why why is equal to come on to time is one of X, the X Information technology'south not integrate both sides, not a constant of integration. That natural log of why is equal to two side the natural log of X. Let's see take exponential is both sides we get why is equal, Teoh abiding a time is next. So the grading of these solutions you by the Ten is equal Teoh a X And so therefore orthogonal trajectories. Nosotros'll take Grady in Do you want? The X is equal to the negative reciprocal of this and so minus two at minus one over to a X. Then integrating this, nosotros're gonna have why is equal to minus 1 over to A. It's only a constant size. The natural log, the absolute value of X plus another constant integration. Let'southward call it D. So now this of has for a given a gives the authority trajectories to the south curves getting here.

So we're given a second gild differential and were given the full general solution to that differential. In that full general solution nosotros have to unknown C. One and C. Two. And so we are going to utilize our given initial atmospheric condition to solve for those. And so starting time we can use that equation and make X equal to cipher and Y equal to one. And we get a 12 equals C. Ane. At present either the cipher is one And eastward. To the zero is 1. Then that cleans up quite quickly. So now let's look at our derivative. Well the derivative East. To the Ten. Is E. To the X. And the derivative of E. To the negative X. Is negative E. To the Ten. Now nosotros can put in the value of ii for our derivative and over again put in that initial condition of zero for X. So again Due east. To the zero and Due east to the negative nothing both equal one. So now find one equation has A plus C. Ii and the other equation has a minus C. 2. So I'm but gonna go ahead and rewrite it and add our two equations together, eliminating C. Two. So nosotros can solve for C. One. So C i equals 3 halves. Given that information we can go ahead and solve for C. Two and it's going to equal negative 1.5 and we're only looking for something. So when we added them together at equal the value of one, then three halves -1/2 equals ane. So at present we tin can write our detail solution every bit three halves E to the 10 minus ane half East. To the negative X.

Were given a second order differential so we're also given its general solution and in that location are two unknowns. We need to solve for C1, ii to arrive a particular solution. So were first given that y nothing equals zero. So that means if nosotros put an X equals zero in the UAE will also be zero. So 10. Either the zero power is i E to the negative zero power is also one. And so very simple equation there. At present, equally we accept our derivatives, the derivative East to the X. Z, the Ten and the derivative of E to the negative X is a negative East to the negative 10. Okay, so now we accept our second situation again, nosotros're going to place zilch in and we get a nil out. So what we find is that zero equals C one minus C two. So let'southward go alee and rewrite that equation right under. Nosotros take a plus C to an a minus C ii so we tin can add them together. Merely what we're going to find is that c one equals zero, and f c 1 equals null. C2 has also equals zero. So the only solution to this differential is y equals null.

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