show-that-the-indicated-solutions-are-in-fact-solutions-of-the-differential-equations-in-the-indic-5

Question

Show that the indicated solutions are, in fact, solutions of the differential equations in the indicated examples.
In Example $3,$ two solutions are shown for the given differential equation. Show that each is a solution.

Vikash Ranjan · Accepted Answer

In the problem the differential equation is voidable as minus wild ass that equals twice halfway. And even solution is why equal power? Sorry This is 484 minus six. Why does is equal to minus four? 84 minus six? While nevertheless is equal to four for minus six. No put it in the equation so it is four for minus six minus of minus four. Football minus six. This is equal to 484 minus six plus 484 minus six is equal to 88 or minus six. Status equal to two into 484 minus six. This is equal to two into well hence we can see that this is a solution to this differential equation so it is the solution to this different cell equation. Hence this is the answer.

Ashley Boni · Answer

so almost will show that this vector X is a solution to d x t t is equal to hey X So we&#x27;re just gonna put this solution and and see that both sides are equal. So first, let&#x27;s take the derivative So d x d t love chance Hide, uh, first wouldn&#x27;t take the derivative of the X component, which gives us on third that we have eight eater. The two tea loves each of the negative t and the drifter But why component to Thursday&#x27;s out front and we have to needed to t and plus e to the negative t So now we need to multiply a time Sex so multiplied knees together we get three times. 1/3 just gives us one. So we just have four feet of to t minus eight of the negative t and then we have negative two times 2/3. So we get minus 4/3 needed the two tea minus feet of the negative t so as their first component, not looking at the second row, uh, two times 1/3. It gives us 2/3 times for each of the two tea minus e to the negative T and then we have again minus 4/3 either of the two T minus to the minus T and simplifying this some more. If we combine together, are eating the two teas. We have four feet of the two tea and negative 4/3 eat of the two tea. So four miles 4/3 gives us 8/3 needed the two tea, and then we have negative one plus 4/3 giving us 1/3 either the minus T for a second component. We have 8/3 minus 4/3 giving us 4/3 eat of the two teeth and then negative 2/3 plus 4/3 which is 2/3 either the minus t and this is exactly, uh, what we started with, so this satisfies our equation.

Bobby Barnes · Answer

we want to show that these three our solutions or our differential equation Why is he go to why Prime plus three y is equal to e to the negative X was just check to see if they work out or not. So let&#x27;s first figure out What? Why Prime ms for each of these. So for the 1st 1 there&#x27;s going to be why Prime is just gonna be negative. E to the negative X. Now we can go out and plug everything in to our differential equation right here. So it&#x27;s going to be two times why Primes. Let&#x27;s just multiply this by two and then three times why? And we want to add these two equations together. We&#x27;re gonna get do you buy? You&#x27;re right. It same way they have it too. Why Prime plus three by is equal to three minus to eat the negative X which would just be eaten named Rex so that one checks him. Now what about the 2nd 1? Well, this is going to be why Prime is eager to. So it should have the same derivative as First Minister e to the negative X and now this should be negative rehabs e to the negative x And again we would multiply the top one by and we&#x27;re gonna multiply the bottom one by two. Actually, those two&#x27;s counts out with each other. Now, when we add these equations so on the left side also have the two by price plus three Y, uh, three e to the negative x plus a negative, too. Heat of the night to extract. Still be e to the negative X and then the e to the negative x over to R E to the negative three House X Well, for those will cancel out with each other, so just be there. So that one check so and then lastly, we can find the derivative. Oh, see? So why pry? There&#x27;s going to be equal to e toothy. Negative Thanks. Her negative e to the negative x plus. Now it will be negative. Three, huh? See, he too negative. Three x over, too. And just like before, we can go ahead and multiply our first equation by the re smoke three. But me you could read So 33 and three And the next one we&#x27;re gonna multiply by 22 and two. The twos they&#x27;re cancel out. And then once again, we get to why Prime plus three y is equal to so three e to the negative X minus two e to the negative extra S B e to the negative X and then the last terms here. Those were just cuts out with each other when we add so that one checks out as well. So it looks like all three of those are solutions of our differential equations.

Taylor Shimono · Answer

so to determine whether our function is solution is a solution to the differential equation. We&#x27;re first going to look at what our differential equation requires us to do to our y function. So essentially, we just want to take the first derivative of why. So when we do that, armed E Y d x will be equal to y prime. Um And so the power rule tells us that essentially, we&#x27;re just going to move three to the front here, and we&#x27;re going to reduce that ex explain it by a power one. So it&#x27;ll give us three x squared over three, which we can see. The threes will cancel. So why Prime is equal to X squared. And when we plug that into our differential equation, we get X squared is equal to expert. So that tells us that our function is a solution to this differential

Rebecca Polasek · Answer

If I want to find the general solution, I would need to do the integral because that&#x27;s the anti derivative of this different television. But first I would need to separate the variables so I would need all the exits on one side and then all the wise on the other. So I want to multiply the D X over to the right side. So my goal will be to have all of the wise on the left hand side and then all of the exes on the right hand side. Notice that unfortunately, there are still these wide terms on the right. Well, if I were to factor out that, why? Because it&#x27;s in both terms, I could factor it out in front. Like why? Times three X squared minus two X that would allow me to move it to the shy because now I can just divide it to the other side. So write it as one over why d y equals three X squared minus two x dx. So to get rid of those differentials, we want Teoh do the anti derivative, which is the integral whenever. Why is of the form one over you and that&#x27;s a known anti derivative. That&#x27;s Ln of you. That&#x27;s a natural log of you. So on this left hand side I had the natural log of why. And then I get X cubed minus X squared plus C because there&#x27;s some constant number there that I just don&#x27;t know what it is. It could be zero, or it could be any other constant number. So we put. Plus C is placeholder now to get why by itself, I would need to get rid of this natural log. Ellen and I do that by using E remember, that is the inverse of Ellen. So we have e to the X cubed minus X squared plus c. Well, it was just pretty complicated there. Now, whenever you have exponents that are adding, it means that the basis of the exponents were multiplying like see how X and X are multiplying. That means we add the exponents and we do that all the time. We just don&#x27;t think about it. So when the bases are multiplying, the X wanted to writing. We&#x27;ll hear. My exponents are adding, which means really I hade the bases that were multiplying so that each of the X cubed minus X squared. I&#x27;m gonna leave it separate because it&#x27;s variables, and I don&#x27;t know what it is, but this e to the C is actually just some constant number. It&#x27;s a constant number because C is constant and e is constant rubber ease around 2.7. So we&#x27;re gonna replace that constant number here. This whole thing with just a capital C. I can do that because of the fact that that whole thing is a constant number. There&#x27;s no variable terms there. So I would actually write my general solution as y equals C E to the X cubed minus X square.

Taylor Shimono · Answer

So since our differential equation involves a y prime right here our first step in solving. But this problem is going to be finding the derivative of why first derivative of why. So when we do that, um, why prime is going to be first equal to the derivative of each of the three x power which we know that we&#x27;re going to take Ah, use The chain will hear. So it&#x27;ll be the derivative of a three X right here, which is three times each of the three x power, which is just the regular derivative of each of the ex power. Um, and finally, it will be minus the derivative of the second portion right here, which is actually just going to say the same You to the X over, too. And the reason that stays the same is just because of the fact that the derivative of each of the ex is e to the X, and so our constant won&#x27;t affect it. And now, in order to test whether this is a solution to our differential equation, I&#x27;ll have to do is set this equal to ah three. Why plus each of the X and make sure both sides are equal. Um, so we are going to substitute in. Why? For this three y on the right side of her equation. And this is going to give us, um, three times e to the three x power, uh, minus three e to the X over too. Plus e to the X. So in this step right here, all I did was distribute my three to my Why that were given on the left side here. And so, in order to simplify, I&#x27;m going to just add e to the X over two to both sides here. Shouldn&#x27;t shouldn&#x27;t have crossed that off there. But all right, plus e to the X over too. Plus e to the X over too. So we can see that our equation is going to be three e to the three. X is equal to three e to the three. X minus. Ah, this is going to be to each of the X over two. So that&#x27;ll just be minus e to the X over too. Or excuse me minus e to the X plus e to the X. And so since these two terms are going to cancel, we get three. Each of the three X is equal to three each of the three X and since we can see that this is true, that tells us that our white is a solution.

Problem

Show that the indicated solutions are, in fact, s…

Problem 1 Easy Difficulty

Show that the indicated solutions are, in fact, solutions of the differential equations in the indicated examples.
In Example $3,$ two solutions are shown for the given differential equation. Show that each is a solution.

Answer

$$\begin{array}{l}
\left(c_{1} e^{-x}+4 c_{2} e^{2 x}\right)-\left(-c_{1} e^{-x}+2 c_{2} e^{2 x}\right)=2\left(c_{1} e^{-x}+c_{2} e^{2 x}\right) \\
\left(4 e^{-x}\right)-\left(-4 e^{-x}\right)=2\left(4 e^{-x}\right)
\end{array}$$

Related Courses

Basic Technical Mathematics with Calculus

Related Topics

Discussion

Top Calculus 2 / BC Educators

Catherine R.

Heather Z.

Kristen K.

Joseph L.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 31

Video Transcript

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Catherine R.

Heather Z.

Kristen K.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$$\begin{array}{l}\left(c_{1} e^{-x}+4 c_{2} e^{2 x}\right)-\left(-c_{1} e^{-x}+2 c_{2} e^{2 x}\right)=2\left(c_{1} e^{-x}+c_{2} e^{2 x}\right) \\\left(4 e^{-x}\right)-\left(-4 e^{-x}\right)=2\left(4 e^{-x}\right)\end{array}$$

Basic Technical Mathematics with Calculus

Catherine R.

Heather Z.

Kristen K.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

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

Catherine R.

Heather Z.

Kristen K.

Joseph L.

$$\begin{array}{l}
\left(c_{1} e^{-x}+4 c_{2} e^{2 x}\right)-\left(-c_{1} e^{-x}+2 c_{2} e^{2 x}\right)=2\left(c_{1} e^{-x}+c_{2} e^{2 x}\right) \\
\left(4 e^{-x}\right)-\left(-4 e^{-x}\right)=2\left(4 e^{-x}\right)
\end{array}$$

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus