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State the type of each equation: Separable, Linear, Exact, SIF , Bernoulli, Homogeneous, or G(ax+by) If the equation is a SIF find the special integrating factor an...

Question

State the type of each equation: Separable, Linear, Exact, SIF , Bernoulli, Homogeneous, or G(ax+by) If the equation is a SIF find the special integrating factor and show that this will make the equation exact STOP at this point: Do not solve. Unless otherwise stated, give the implicit form of the solution.(3)I+cos" (t-x) =(4) (2xy-3x" )dx+(x" - 2y ')dy=0, y(0) =4

State the type of each equation: Separable, Linear, Exact, SIF , Bernoulli, Homogeneous, or G(ax+by) If the equation is a SIF find the special integrating factor and show that this will make the equation exact STOP at this point: Do not solve. Unless otherwise stated, give the implicit form of the solution. (3) I+cos" (t-x) = (4) (2xy-3x" )dx+(x" - 2y ')dy=0, y(0) =4



Answers

Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(x)=y \cos x, y(0)=3$$

Take a look at the differential equation. D Y over the X equals four e to the negative three. Exit the differential equation because I have that differential d y over the X on the left hand side. And if I want to get back to the general solution, then I want to give her the do Iraqi ex. I want to go back to, like a y equals equation. So in order to that, multiply your DX to the other side. This separates are are variables for us. Notice how now? I would just have the wise on the left and the X is on the right in order to eliminate thes derivative these differentials I want to integrate, which is the anti derivative. So the integral of d y. It's like one d y is why and then over here on the right side, this may not be the easiest to just take a look at, but it's a great time to try and practise you substitution. So if we replaced the exponents with the negative three x, the derivative of that would be negative. Three d x and I only have the DX of that in my integral. So I like to solve for what I have, I have DX so rewriting this right hand side would be for E to the u times. D you over negative three. I replaced the negative three x and I replaced the DX with the two new terms that I wrote out on the right hand side. This makes it easier to solve because that negative 4/3 can come out front as a coefficient and e to the U in the interval has an anti derivative of E to the U Just make sure for your final answer that you take it back to the original variable. So want to get back to the negative three x as my exponents and then I have to write plus C because I don't know what constant number could have been there at the end. And that plus C is a big part of what makes this the general solution because I'm going to

Let's solve this problem by solving for the homogeneous solution. So substitute out the Y double prime, the Y. Prime. And the white terms up with our square plus four R plus three equals to zero. We can factors out so we'll get that are plus one Times are plus three equals to zero. And this gives us our values negative three A negative one. So with that we can actually build our homogeneous solution is homogeneous solutions can be C one E. To the negative X. Policy to eat in a. of three x. And with that we can actually take a guess at our particular solution. So our guests for the particular assertion Y F P is going to be in the form A. E To the -3 x. Now note that there is a term here in the homogeneous solution that matches up with something in a particular solution. And since we don't want that, what we end up doing is we all end up multiplying by factor backs for our guests with a particular solution. And so now with that we can actually take the derivative of this twice, you know Y. P. Prime. It's gonna be E. So negative three X. With the product rule minus three A. X. E. 2.5 3 X. And Y. P. Double prime. It's gonna be negative three A. E. To the negative three X minus three A. E. To the negative three X. Plus nine A. X. E. to the -3 x. And so now let's simplify this this Y. P. Double prime term. And so this can condense into negative six A. Eat and I have three x. Plus nine A. X. Eat the May have three x. And so you can plug this into our original equation. And so original equation was Y double prime. So negative six A beats the knife three X. Plus nine A. Ex eats an F. Three X. Plus four Y. Prime. So plus four times a feat than they have three X minus 12 A. X. E. To the negative three X. Plus three Y. So plus three a. It's an area of three x. x. And this equals to eat the neck of three X. And so let's try and simplify this down before we build the systems of equations here. So are like terms you're going to eat this, this and this and the other. Like terms are gonna be this this oh sorry um And this place the X. In the wrong position there. And so now we can connect these terms or simplify them down. So have negative to a Eat. And they have three x. Plus well this plus this will cancel it against this. So this will be They have to a seat and a three x equals two Eating i. e. to the -3 X. Now we can divide by each of the 93x. On both sides. So cancel out. So I have that too. Two equals one Or that a equals negative 1/2. And so with this we can actually build our particular solution. Our sorry our total solution. So our total solution is a summer for homogeneous solution and our particular solution. And so are homogeneous solution in this case was C one Ethan negative X Plus c. two e. 2. And they have three x. And we just got our particular decision which is negative one half X. E. To the negative three X. And so this is actually her answer.

For this equation, we need to find solutions of the form wise equal to eat the Rx. So next we need to find why Prime, which is gonna be our either the Rx by a double time, which is equal to R squared either the Rx. And then why Cube, which is equal to our cube either the Rx. So now we substitute these into our differential equation here. So we get our cube. You know, there are X plus three R squared. Either the r X minus four are either the Rx minus 12 e to the R. X, and that's equal to zero next week. In fact, out in either the Rx from all four terms here. So we get either the Rx and then what's left overs are a cube plus three R squared minus four. R minus 12 is equal to zero. Since either the Rx can never be equal to zero, we need to set this part of the equation included. Zero. That's our cue. Plus tree R squared minus four. R minus 12 is equal to zero. Since this, um, constant coefficient here. Is it with one by the rational root beer? Um, we only need to look at the factors of 12. So includes plus or minus one plus or minus 12 plus or minus to plus or minus six plus or minus three and poster minus four. Right. So we're going to go and try to find the roots of this by using synthetic division. So 13 negative four and negative 12. So we contest which of these are going to be roots. Okay, so I want to start with one to see if one is a root of this equation. So one. So I bring down the 1st 1 then one times one is one. Then I add those to you. I get four. Um, then one times four is four. That becomes 00 and negative. 12. Since I don't have a zero here, one is not a route. So I'm gonna start over, so erase those. So now I'm gonna try a negative one. Okay, so let's try. Negative one. One times. Negative One is negative. One that becomes three plus negative. One is to hear one times negative. Two is negative. Two is becomes negative. Six, which comes positive. 12. Um, sorry. Um, this becomes positive. Six and again it's not. Ah, so this is not a route as well. Okay, um, so now let's try to okay, Race nous. So let's try to here so to I'm gonna try to times one is two. They gives five here. 10 um, so 10 minus four is six than two times six is 12 0 So two is a root of this equation here. So if we factor that we get our minus two and then are left over coefficients and we are square plus five r plus six is equal to zero. So now we can factor this further. The factors of six at up to five are two and three, so that's gonna factor into our plus two. And r plus three is equal to zero. Therefore, our roots are gonna be r equals too negative too. And negative three. So we're gonna hav e to the two x e to the negative two x and e to the negative three x as are linearly independent solutions Now our general solution. Why Sea of X is going to be equal to C one e to the two x plus c two e to the negative two x plus c three e to the negative three x

All right, so we have depression. Pollution, white time affects close white costing X first with the right white prime of excess. B Y over B X people. So why call Scientifics? Hey, And as we can observe, this can be, uh, sold by using the separation of horrible method. So we divide both sides by y and then deploy boat states paid the extra. Have you? Why over wide? Because causing the fax? Yes, me. And then we get the unsavory sites. We haven't developed the way over. Why? Which is Ellen off. The absolute value of flying equals two legal. Of course saying xcx, we have sign of X. And don't forget the arbitrary constants C Okay, So, in order to sell for why get the eat powers boots sites we have here east Eleanor. Why just wide equals your isa sine X and then listen or the skin breathe an ass. Here is the sine X. Okay? And we're us to get the particular solution. Since we have initial value, Lucious, while zero coast. Okay, so this is a X equals zero. Why is it called? Okay, there. So we have three. Because here is a sign of zero times. Hey, and have the equals being reached Asylum 00 And this year is zero for two r and then we have here is to see therefore you re Stasi's off the city. Okay, So substituting this value to our ah general solution. Here we have Oy, because here is the sine X eight times 34. We have the You're Easton. Sign off. It's Hey! So this is over solution.


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