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(3) Solve the given system of 'differential equations: x=X +2x2 ; X1(0) =-1,and x2 =4X1 +3x2 ; X2(0) = 4...

Question

(3) Solve the given system of 'differential equations: x=X +2x2 ; X1(0) =-1,and x2 =4X1 +3x2 ; X2(0) = 4

(3) Solve the given system of 'differential equations: x=X +2x2 ; X1(0) =-1,and x2 =4X1 +3x2 ; X2(0) = 4



Answers

For the following problems, find the general solution to the differential equations.
$$
y^{\prime}=3 x-2 y
$$

Here we have the differential equation ex. Dont expose three y equals four x squared with an initial value or value at least of why of five eagles 22 and so to use our integrating factor method What we first need to d'oh is, um get this in the right form. So we need to divide everything by axe to get do I d. X by itself. So get do I d x plus three over x times Why equals for us And now we can find our integrating factor by taking e to the integral of three over x t x, which is e to the three natural log of X, which is X cubed right before we can cancel the e in the natural log we have to bring the three inside said it's eat the natural log of x cubed and then they cancel out. So we're gonna do is multiply everything by X cubed And so this is what we'll get And now remember the trick That's really nice about using our integrating factor so you can recognize this thing on the left is the derivative of X cubes times why, uh and so now That means that X cubed times why is integral with respect to acts of this expression on the right, which is just X to the fifth plus c. Okay, so that now, um what kind of come back over here and just to get why we're not divided by ex Cubes. So why will be X squared plus c over execute? Okay, so the last thing we want to d'oh huh is we want to use our initial value or the value that we have to solve for C. So when we plug in X equals five, we should get 22. This is 25. We'll see over 1 25 Mrs. C is gonna be 1 25 times negative three the way we first subtract 25 then multiply by 1 25 which is minus 3 75 Okay, so you can just substitute this back in as our final expression, and there we go

To solve this differential equation, we first need to put in the right forms. We can use our integrating factor. That is, we need thio get you ready x by itself. So she did divide everything by X, Okay, And then remember integrating factor. We confined by taking e to the integral of the expression in front of the white term. So we can first rearrange what's inside the integral. It's one over X plus one if you just distribute in the bottom here. So this is gonna be e to the natural log of X plus X, which we can simplify his axe times in Vieques. Okay, send a silver equation. What we're gonna do is we're gonna multiply everything by this integrating factor. Swine goals three over X times X of the X. Okay. And then what always happens in these cases with all the computations are done right? Is that the thing on the left is just the derivative of integrating factor times. Why? So then we just had this equal to what's on the right hand side that you can supervise a little bit to be three e. T. X. So then, if we just integrate both sides. We get the backseat of the X Times. Why should be the integral of the right hand side? But since this is just three envy X, we get that back. It's a constant and then solving for why we get three over X plus c over X times e to the minus X is the general form the solution? But we still just need to use the point that we have the one access for y equals 50 to solve for a constancy. So 50 more vehicle to the right hand side with four substituted in for X everywhere. It's okay, So then to solve for C, we just want to move everything else over. I believe that we should get so he would subtract 3.4, which would give us 49.75 number multiplied by four and multiply by E to the fourth. And this should be 1 97 beauties of the fourth. It's okay, so then we can substitute the sin. So we get wise three over X and for C. We're going to substitute this in, so we're gonna get 1 97 over X and then we have this seat of the fourth, but then weaken, group it in with the even minus X like this as eat of the fourth Mind sex. So this is the solution to a different feel equation.

Right. So we have a system of three. Ah, first order differential equations. And, um, at first you might think a little bit hard because it's a three by three system. But then we'll realize that, um, our first equation is actually completely independent of the other two. Notice how it's only dependant on the next one and X neither x two or Norex three are dependent on X one. So since this is independent, so this is independent of X, who next? Three, we can go ahead and already right it solution by just knowing that this is a form of, um, exponential function. So now since we have that, we'll have to go ahead and solve do this system. So ah, if we rewrite an operator form, we'll end up with, um, Do you mind us one x two plus sex three zero and ah, minus X two plus D minus one x three is equal to zero. So then we'll go ahead and perform the D minus one operation on our top equation again. Then the D minus one squared X two us. Do you mind? This one extra t is equal to zero. And, um, by moving this to the other side. We know that our D minus for next three is just x two. So we'll get that. Do u minus one squared x two, um, plus X to the zero. So then we'll have to expand our equation to get something of this form so D squared. Let's two d plus two x two is You could have zero. So now, um, we'll go ahead and solve this, um, with the coordinated formula. Um, sorry. It was a little bit healthy, but And the squared minus two lambda plus two to quote a zero. And we know that our Lando's Oh are, um, two plus or minus the square root of four minus a all over too. So this gives us, um so this is the square root turns into minus spurred for I? No. And then, um, that just gives us, uh to I. So we have two plus or minus two I over two, which is just one plus or minus. I So then, since our Lambda Zehr one and pleasure my side, we know that our X two should equal see to you to the t co sign t. Let's see three. It's the tea side of T. So now we have to use our relationship from our first equation that, uh, x three just ah negative. D minus one x two. So now that we know that we can go ahead and and, um, find what this is, so then we'll say that X three is just equal to negative. Um And then we have to use our product through here, so negative will take the different that the derivative of X one of X two events attract one next to So this gives us, um, see to e 30 co sign T um, minus C to each t scientist, E plus C three eat t signed t Let's say three. You 30. Ko Sai Inti. I didn't attracting one multiple of both. That gives us minus C two. It's a ti ko 70 and minus C three. Either. T scientist, we have a lot of terms that we can go ahead and combine here. So we have go ahead. And Markham Mark all of his e t co seventies and red. So you you didn t co sign you to Vico Sign it because I and then are signs so we know. Decide, um, this term in this term, cancel. And this term in this term, cancel. So then we're just left with you. Ex three is equal to negative. Negative, uh, see, to read to the coast and so see, to be to the sci fi And then minus C three, it's the tea co sign t. So this is our final form of X three. And here's X two. So, um, these are our the re solutions.

So find the solution of this differential equation. First, we need to put it in the right form so that we can use our integrating factor. So we'll need to do for that is, divide everything by two and then move this five x turn over to the other side. So if we do that, we'll get D Y t X minus five X y equals five halfs necks. That's now this isn't the right form. So we confined our integrating factor by, um, setting it equal to e to the integral of minus five x just dysfunction that is in front of the white term on the left. And so we can we can just take this integral. It's just ah, our role So e to the minus 5/2 x squared. So this is what we're gonna multiply our equation by. So get e to the minus 5/2 X squared times the White House minus five x e to the minus five halfs act squared. Why equals that house x times e to the minus five halfs X squared. And now remember, a trick is that this term on the left should be the derivative of are integrating factor times. Why? And you can check that this is right. Just buy the product rule. We're not gonna do anything to the right hand side just yet, but its expression tells us is that if the we have a derivative of this expression on the left is what's on the right, then we can integrate both sides and eat of the minus five house X squared times. Why gonna be be integral of this expression over here? And, um, this is just about, uh, going toe all cancel out nicely. We're gonna get some constant times e to the minus 5/2 x squared, and I believe what's still gonna sit out front is a minus 1/2. If we want to play down by the derivative of the top here, we're gonna multiply by minus five, Tex. And so if we take it we have here, we'll get 5/2 x times his expression. Yeah, I think this is right. Plus C and then remembered t find, uh, why itself? We'll just divide by what we have, and we'll get minus 1/2 plus c times e to the five halfs X squared


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