show-that-the-given-equation-is-a-solution-of-the-given-differential-equation-x-yprime2-y-quad-yc--3

Question

Show that the given equation is a solution of the given differential equation.
$$x y^{\prime}=2 y, \quad y=c x^{2}$$

Vikash Ranjan · Accepted Answer

In the problem we have X wide as that is equal to Y. So here the solution given is Y is equal to see X Squire, Or my last is equal to see into 2X. That is equal to twice X. C. So they put this in the decider equation, so X into two X. C. That is equally to into Y e c x Squire. So it is too Squire. See that is equal to two x Squire. See, and we can see that why is equal to see X squared? He&#x27;s a solution.

M S · Answer

What&#x27;s the show? That this is a solution to our deferentially, please. So, first, we&#x27;re going to take the derivative of dysfunction here. So why Prime equals two birds ender got a movie to the exits. Just needs to be it now for a second. Luke. Clues derivative. Ah, eats the negative. Two x is a derivative of this here, which is negative too. All times e to the negative two weeks so we can rewrite this as 2/3 need to be X minus two times e ds and negative to it. Now we need to somehow show that this equation and it&#x27;s equal to this equation. But first, we need to get rid of this y variable here by plugging in our original Why equation that was given to us. So we&#x27;re going to say 1st 1 to subtract two are from both sides. So have negative too. Why plus two times each Be X now we&#x27;re gonna play gin are y equation. So why pride equals negative, too. Times 2/3 e to the x plus e to the negative to x all plus two times e to the X. Now British retards who will have negative 4/3 usually X minus two times E to the negative two ex all place two times in two New Yorks. Now, this is almost similar to this here, which is what we want. However, we can combine like terms because these two have the same base here. So too, if we give it a denominator of three, our top will become six because an original denominator is one what&#x27;s part of top and bottom by three. So we get 6/3 I&#x27;m gonna be right there. Says 6/3 here and now we can combine thes two times here. So we have 6/3 minus 4/3. That gives us positive 2/3 into the X and then bring down our two times eats of negative to X. And because this here is equal to this here, we know that it is a solution.

Alexandra Embry · Answer

in this problem were given a differential equation, and we want to see if this is an implicit solution to this differential equation. So to do that, let&#x27;s take the, um, derivative of this equation using implicit differentiation. So we&#x27;re gonna use the product rule here and derivative of excess one times. Why Squared plus X times. The derivative of Y squared is two y times D Y DX, plus to times D Y. D X minus one equals the derivative of CIA. Constant is zero. Now we want to solve four d y t x, so to do that, let&#x27;s move this y squared and negative one over to the other side and pull Factor out d y d x from what we have left. So we get d y dx times X times two. Why plus two is equal to one minus y squared. Then to get do I d. X by itself weaken Divide each side by X times two y plus two and we get D Y D X is equal to one minus y squared over to We can factor this two out of this entire thing to make it to X y less one and we see that D Y t x or Y prime is equal to one minus y squared over two times the quantity one plus X y. And so, yes, this equation does solve our differential equation.

Ernest Castorena · Answer

They asked us to verify that it satisfies the central vision for Let&#x27;s Find the first year all of negatives he wanted for the negative to be for the X. We take our second riveted and we&#x27;ll have C one e to the negative. We&#x27;ll see to be to the X notices. The same thing someone has attracted him. One from me of it well, under 10 does satisfy the differential equation.

Alexandra Embry · Answer

in this problem. We are given a differential equation over here and an equation over here, and we want to decide if this is a solution. Teoh this differential equation. So to do that, we should implicitly differentiate this equation over here and see if we can get um ah, why Prime Isolated. That looks like this equation. So we can go ahead and do that. I&#x27;ll do it over here and blue. So eat the Y over X. We have to implicitly differentiate. So we&#x27;re gonna use the chain rule here so e to the Y over X and then we&#x27;re gonna multiply by the derivative of Y over X, which is it&#x27;s a product role because it&#x27;s why times one over X. So we&#x27;ll start with driven of of Why is D y d x times one over x plus why times the derivative of one over X is negative and X to the negative two or one over x squared. So that&#x27;s our first term plus product rule again differentiating this one times Why squared plus x times two d y. Oh, sorry. It&#x27;s gonna be X times two Why it&#x27;s using the power well here too wide to the one times D Y d X minus one equals zero is the derivative of our constancy. So now that we have this, we want Teoh isolate these d Y d exes And so let&#x27;s move everything else that we can to the other side and distribute out this e to the y over X So we get e to the y over x times de y d x times one over x And then, um, we have this over here this x times two i times do I d. X on the other side We&#x27;re gonna add this one and then we&#x27;re going to subtract this white squared and we&#x27;re going Teoh, add this term over here with this e to the Y over X distributed to it. So that will be plus E to the Y over x times. Why Times X to the negative, too. We&#x27;ll cry then on this side. We can factor out the d y d x from both of these terms and then divide both sides by the inside So we get d y DX equals one minus y squared plus e to the y over x times Why Times X to the minus two all over this when we factor out the d y DX So all over e to the y over x times one over x plus x times two Why? So we&#x27;re getting close to the answer that we want. Um, so let&#x27;s see if we can make this de y t x look a little bit more like this white prime just to confirm that we have the right thing. So up top, we can Let&#x27;s see. Okay. In the bottom, they have an X factor it out. Let&#x27;s go ahead and rewrite this with our X is factored out. Best we can, um, here. So on the top we still have one minus y squared plus e to the y over x times y times extra minus two The bottom We have e to the Y over x times. Let&#x27;s factor out this one over X so well, this will just be times one plus x squared times to why all of this times one over X. You see, when we factor this room, we distribute this one over. Expect in here. That&#x27;s we get the same denominator. Now we have this one over X in the denominator. So that&#x27;s like multiplying by X at the top of the numerator. So let&#x27;s do that. This is equal to We can go ahead and scroll down a little bit. Over here we have one minus. Why squared Times X Because we&#x27;re multiplying everything in the top by ex getting rid of this one over X in the bottom plus e to the y over x times Why Times X to the minus one all over e to the y over X plus two X squared. Why now? Let&#x27;s look back at what we want this to look like we&#x27;re close. It looks like we should just factor out a one over X at in the top as well. So we get we factor out X to the minus one In the top we get one minus y squared times X squared plus e to the y over x times y all over e to the Y over X plus two X squared. Why? And then we see this X to the minus one. We can just bring down to the bottom to get our final answer of X squared one minus y squared plus E to the y over x times y all over x times e to the y over X plus Teoh X squared. Why? And then we see that, Yes, this equation over here is a valid solution to this differential equation.

Samuel Hannah · Answer

to solve this differential equation. We first want to get it in a form that&#x27;s familiar to us of D y over the X by rewrote rearranging the question receive it D Y over the apse is equal to y over x times X squared minus Wadsworth, which is divided by ex word plus was word We&#x27;ve got to simplify this on. We can do it by looking at the second fraction So we leave Why of racks It is And then for the numerator We fact around ex word and then that becomes X squared multiplied by one minus one square over X wet. We do the same thing for this nominated time that becomes X squared After we pulled the factors in front multiplied by warm plus voice word over X split. We don&#x27;t have the common factors of X squared in both the numerator on the dumb nominator. So these canceled and we must see that this is a homogeneous differential equation because it is dependent only on why Who for X As such, we can make the substitution of B equals y of Rex equals. Why of the X Andi This gives is that the left hand side becomes the Times one minus fees. Glad over one. Plus he&#x27;s what to the left hand side. Need to rewrite the wind. The X So if we let why equals the Times X, then the Juan de X by the product rule becomes B plus x d plus x devi. The X, which when we put on here, gives us a differential equation in terms of the Now, if we take this the on the left hand side, over to the right hand side. By subtracting both sides by the we can rewrite the equation on the right hand side to a more to think manner. So free combined the two terms by multiplying the minus feet. By one plus B squared, this becomes a fraction. The times one minus V squared, minus the times one plus B squared well ever one plus V squad on when we spend at the blankets and simplify that becomes minus to the cube over one. Plus, he&#x27;s glad, and then they get on the left hand side of X DVD X. So now these two terms the first and third terms form a separable different liquidation. If we separate this by bringing the B times to the left and the extent to the right. We see that one plus B squared over to be cubed. TV equals minus one over X, the EPS and then to get a solution for the we need to integrate if sides now we will. The impression of minus one of Rex is fairly simple, cause it&#x27;s just being negative of the natural log of X. So we will leave that for now and focus on the integration of the V parts. So if we write this out the integral of one plus fee squad over to V cubed you think be written as ah Hoff begin to grow as one of the the Cube plus one over the TV. We have taken the fact the constant factor of 1/2 outside the integral, then simplified the fraction by splitting into the two fractional components. Now we know V integral is linear. So we integrate one of Avi Cube, then one over the on. These are fairly simple to do and this gives is the overall integral is move times the natural log of e minus on over for the squad. So if we now combine this with the earlier results. We see that the solution the perceptible equation is oh, natural log the fee the absolute. Both in there, minus 1/4 V sward equal Tu minus natural log. Lex plus some integration Constant. You know, we can simplify this. I bringing the natural log comes perform TV raft, onside and the other terms that have right. So this consistent half l envy us l a next equals C plus 1/4 B squad and then using the natural longer of most we can combine the row comes on the left hand side to give us Ellen E. X squared over to equals. See bus one over for the squad. Now, this is no us, no solution solely in terms of E. But it&#x27;s an equation which satisfies all differential equation. And it&#x27;s in a neat formula services the solution we will take now all of its remaining is to get a solution for why, on we need to remember, but B equals y of racks. So if we substitute this in, we get it not to lug of the x glad becomes the natural log of y times x at the two and that&#x27;s equal to Semenza crimes, installation costs, etc. Plus X squared over four y squared on. This is the solution, but we take we&#x27;re different. Jill Equation.

Problem

Show that the given equation is a solution of the…

Problem 12 Easy Difficulty

Show that the given equation is a solution of the given differential equation.
$$x y^{\prime}=2 y, \quad y=c x^{2}$$

Answer

$y=c x^{2}$ is a solution

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$y=c x^{2}$ is a solution

Basic Technical Mathematics with Calculus

Catherine R.

Heather Z.

Caleb E.

Kristen K.

Integration Review - Intro

Definite vs Indefinite

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Catherine R.

Heather Z.

Caleb E.

Kristen K.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus