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Solve the given initial value problem.dx 3t _ 3x+y - dt dy =X+3y; dtx(o) =1y(O) = - 3The solution is x(t)and y(t)...

Question

Solve the given initial value problem.dx 3t _ 3x+y - dt dy =X+3y; dtx(o) =1y(O) = - 3The solution is x(t)and y(t)

Solve the given initial value problem. dx 3t _ 3x+y - dt dy =X+3y; dt x(o) =1 y(O) = - 3 The solution is x(t) and y(t)



Answers

Find the solution to the initial-value problem.
$$\frac{d y}{d t}=y^{2} e^{x} \sin (3 x), y(0)=1$$

To solve this initial value problem. The first thing they could do is divide each turn by CI Cube to get us into the standard form. Why? Prime plus p of x y is q of axe. So again, I'm dividing each term. Okay, Integrating factor. Each of the integral p of t d t. We know this integrates to eat the three natural log of t which is the same thing as eat the natural log of t cubed. Remember, each the not walk of is simply one which means we have t cubed is our integrating factor multiplying the differential equation by the intrigue factor In other words, each of the terms we end up with us now integrating both sides we can solve for why soul For why, by dividing each of these terms by t cubed order to get why is teach the negative three sign of tea costs each e to the negative three Now use the initial value. Why of pi zero In order find See we have zero is pied to the 93. Just think of this accident. Why, that's literally what we're doing. We're just plugging in our accent. Why values because again we're trying to solve for C, which is our constant of integration. We get zero is equivalent to see because 00 plus c pied to the negative three. So see a zero, which means our solution as why, as a sign of tea divided by t cubed.

Alright so here we have t cubed Y. Prime plus four T squared Y. Is equal to E. To native team. Were given that wide native land deals here. So our first goal is to get this white crime alone. We're gonna have Y. Crime plus four over T. Y. Equal to pete negativity over TQ. Yeah so um now assigned P. O. T. To be for over tea. F. F. T. Is equal to the integral of four over T. T. T. It's illegal to for natural log of T. So a multiplying factor is going to be E. To the fore L. N. T. Which is just equal to T. To the fourth. Alright multiply that all out T. To the fourth. Y. Prime Plus four T. to the 3rd y. Is equal to T. E. To the negative teeth. Okay. Okay so now we're gonna make our product on the left hand side is the D. U. D. T. Of T. To the fourth Y. To go to T. E. Two negative T. So we integrate both sides respected to you some D. D. T. T. 4th Y. T. T. You go to the integral of T. E. Negative T. D. T. The usual cancel we're gonna be left with the integral of T. To the fourth Dy. Which is going to come out to the T. To the fourth. Y. Is equal to. And then the integral of um uh sorry the integral of T. Times E. To the T. Is going to be um negative T. E. Two negative T uh minus is related T. Plus E. So now we're gonna get the wild jones Y. Is equal to we're gonna have negative E. To negativity over T cubed minus E. Negative T. Over to the 4th classy. Yeah so now we are gonna call in our initial condition Which is why you why have negative one equals zero. So we're gonna physical 20 is equal to eat a negative or I guess negative one is going to eat the one um over and then negative one cube is still negative one minus E. To the one and negative Q. Negative one to the fourth is one plus. See So now we have so these are gonna have a lot of positive. Now let's just eat the 1/1 minus. Is the one. Number one plus C. Goes there these cancel and C equals zero. Or final equation is just going to be Y equals negative ease and negative teeth over t cubed my as E. To the negative T. All over 2 to the 4th. Okay.

The first step here is to take the loss transform of both sides in this equation. And when we do so we're gonna get the loss transform of why prime plus four times of the loss Transform of why, plus being equal to three times with boss transform of Delta of T minus one and using the rules that we have up here with the class transforms. We can rewrite this as s times little boss transform of why minus lion, zero plus four times of loss transform with why being able to three times e to the negative s where a is one in this case. And now we can solve capital. Why they factoring it out? So I get why times as plus four minus y evaluated zero is too the angle that's the region e times any of the negative s and selling for by capital why we're going to get three e to the negative s plus two proud of I s plus four. We can write this as to two separate fractions e three e to the negative s over s plus four plus two overexpose For now, we're gonna take the inverse laplace transform on both sides The inverse LaPlace transform of the boss transform of why we'll just give us why Being able to converse a plus of three inch of the negative s over as plus four plus in versatile class of two over s plus four. Now we can look at each of these individually universal falls transform of three into the negative Yes, times one over X plus four. We can take the three outside of the operation using the linearity of little Bosh transform. And we can use the second shifting through room to rewrite this as three times the heavy side function at one of tea. It's our A Up here is one times the inverse LaPlace transform er one over X plus four, which will be e to the negative four times T. We have to subtract the base of our every side function some minus one and looking at the inverse of loss transform of two over s plus four. You can take the two outside using the linearity of the boss transform and we know, but the in vessel class of one of us plus four Well, the two times e to the negative for teen since is negative for been similarly for the last one. And now we can write our general solution as three comes the heavy side at one of tee times E to the negative four T minus one plus two times into the negative 14.

The first step to our solution will be taking the applause transform of both sides of our equation. Such that will get the boss transform on the second derivative minus three times of the lost transforming the first derivative plus two times of boss transform of why single to the class transform on Delta of T minus one. Now we can use what we know of the class transforms to simplify this equation. So we're gonna get s squared times of the blast transformer. Why? Minus s times Y zero minus y prime. And zero minus three times s times wireless minus y of zero was two times Capital Iris being able to little boss transform of Delta T minus one. Well, a is one in this case, was, will you to the negative s. Now we can gather all of our terms here to get as squared minus three s plus two all times capital. Why minus why have zero times s and r y of zero was one so minus s minus y primer and zero, which is zero minus three times negative one. So be positive. Sorry. Being able to you to the native s now we can solve for our capital. Why which will be equal to you need to the negative s plus s minus three, all divided by s squared. Minus three s plus two. And we, in fact what we have in the bottom in the denominator to be you need to Negative s plus s minus three. All divided by s minus two times asked my response. Now we can separate this into two separate fractions yet you need to the negative s not by s minus two. Tens of slyness. One plus s minus three. All invited by s minus two times X minus one. And now we're gonna partial fraction decomposed. Both of these, starting with the fraction on the left. Really get one over X minus two times X minus one Unical to a over X minus two plus being over X minus one. Multiply the denominator on both sides to get to This is one music with you. A times as minus one plus B times X minus two benefits. It s You were the one. We're going together. This is one is equal. That a time zero plus de times one minus two, which is negative. One therefore be tested people. The negative one never said s equal to to really it. That one is equal to a times two minus one, which is one plus any time. Zero. So therefore, a has to meet with one. It's therefore, this is one over. X minus two minus one over X minus one. And now we're gonna look at the other fraction that we had, which was s minus three. All divided by s minus two times X minus one. We want to separate this into another fraction. A over s minus two plus being over s minus one again. Multiply the denominator through together. This is s minus three is equal to a times X minus one plus being times X minus two And again, if we said ask being eagle, the one we're going to get one minus three is negative. Two on the left you mingle, Teoh. A time. Zero plus being times one minus two, which is negative one. So therefore, being passed a vehicle to to never said s being able to to really a to minus three, which is negative. One on the left times A times two minus one is one plus being time zero. So therefore, a has to be equal to negative one. So therefore, this is going to be negative. One over X minus two plus two over X minus one. Now we can take all this, plug it back in for a capital. Why? Expression, which will be our first fraction, which will be eating the negative s turned all of this. So using the negative s times one over s minus one minus each of the negative s times won't ever asked. Minus two. The answer is backwards as minus two is positive in this minus one is negative. Plus the second fraction, which is negative. One of Rs minus two plus one plus two over s minus one. Now we're gonna take the inverse applause transform of everything to get back our Why. So we'll have the inverse laplace transform Negative musical negative s times one of us minus one plus in versatile plus of means of negative s times won't ever asked. Minus two minus in verse the plus of one over X minus two plus two times in verse, The plus of one over X minus one. And now we can look at the inverse or fact that we know from our rules for applause transforms. That's a plus. Transformer one over s minus A in Bristol class of one of rest mind say, is key to the 80 here is too. So this is just e to the negative to teach. And this was the to be a positive too. And this will be easy to the positive t. And we can use the second shift serum to write this. Since we haven't e in our inverse LaPlace transform as negative heavy side function of you raised to negative A and A's won don t times the inverse LaPlace transform of one Arrest minds one which Richard Sermon was e to the T minus the base of the part of the heavy side function wishes one and same thing. Over here. This will be every side of one. I'm t times e to the two times t minus one. Now we can write our general solution as factoring out the heavy side function new one of t times e to the two terms T minus one minus E to the T minus one minus a to the to change plus two into the team


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