5

Consider the following differential equationf+Sy+6y=12e 4 +6t_y(0) = -1 , y(0) = 0The general solution is5 35e-2t +4le-3t +l2e 4 +t_ 6 = 4le-2t -35e-3t -6e 4t +t-5 ...

Question

Consider the following differential equationf+Sy+6y=12e 4 +6t_y(0) = -1 , y(0) = 0The general solution is5 35e-2t +4le-3t +l2e 4 +t_ 6 = 4le-2t -35e-3t -6e 4t +t-5 64le-2t +35e-3t +6e 4t +t_ 5 65 4le2 +35e" +l2e 4+t+- 6

Consider the following differential equation f+Sy+6y=12e 4 +6t_ y(0) = -1 , y(0) = 0 The general solution is 5 35e-2t +4le-3t +l2e 4 +t_ 6 = 4le-2t -35e-3t -6e 4t +t-5 6 4le-2t +35e-3t +6e 4t +t_ 5 6 5 4le2 +35e" +l2e 4+t+- 6



Answers

Find the particular solution of each differential equation for the given conditions. $$D^{2} y-D y-6 y=5-e^{x} ; \quad D y=4 \text { and } y=2 \text { when } x=0$$

We'll start off this problem by solving for the homogeneous solution first. So we'll have r squared minus five. R plus six equals zero. If we deal with the factoring we should get our -2 times AR -3 equals zero. And this gives us our values of two and three. And so now with this we can actually build our modular solutions or modular solution is going to be C one E. To the two X plus C. Two E- three x. And also we can take a stab at our guest for the particular solution. So our guests for the particular solution, Y f P is gonna have to form a. E. To the X. We'll need to take the dirt of this twice. So the first derivative YP prime, it's gonna be A E. To the X. As well. N Y P double prime is also going to be a. E. To the X. And are using these three equations that we just arrived. We're gonna want to plug it into our original equation. So original equation was Y double prime. So E to the x minus five Y. Prime minus five times E. To the X plus six times Y plus six A. E. To the X. It's all equals to two E to the X. Now we can do a little bit of simplification here. So this should simplify down to be to A. E. To the X equals the two E to the X. If we divide by each of the X on both sides will have that to eagles too. And that gives us a value of one. And so a particular solution, answer is going to be each of the X. And we can find our total solution or rather build it by adding the particular solution to the hamada solution. So we're gonna have each the X plus C. One, Each of the two x plus c. two. Each of the three x. Now with this we can actually plug in our initial conditions Which was 50 equals one. And why Prime zero equals to zero. So we're gonna take the drone of Y. F. T. First. So A a Y. T. Prime, It's going to be each of the X. Plus to see one each of the two X plus three C. Two. E. Two the three X. Let's plug in these initial conditions. So I'm gonna plug in zero for X. So we'll have see one Plus C two equals to one. If we plug in zero into prime for where we see an X. We should get that to see one Plus three. C 2 equals to zero. And so with this we can actually build a system equation or we have a system equation. So we'll solve it. And I'm gonna multiply the top by three. So we'll have three C one plus three C. Two. He caused it three. The bottom we're just going to carry it down. So I have to see one plus three C two equals to zero. If we do subtraction we cancel out c. two So have c. one equals 2 three. And with that we can plug it back into original equation. So we say that C one plus C two equals to one. We know that see ones three. So three plus C two equals to one. If we subtract by three on both sides we get C. Two to be equal negative too. And so with this we can build our total solution. That's our final answer. So our final answer is gonna be E. To the X plus C. One. In this case C one is three, So three each of the two x minus two. E. 23 X. And that's your answer.

So start off this problem by substitution with the wide of a prime and the Y prime in the white term. So we'll do our square plus five are plus six equals zero. And this will be these factors to our pledge three plus are plus two close to zero. And this gives us our values of negative three and negative two. And so with this we can actually build our homogeneous solution. So why homogeneous is going to be C one E to the negative three X. We'll see to each of the negative two acts and we can take a guess at our particular solutions. Our guest for a particular solution, it's going to be a E. To the X. And we want to take the do this twice. So we're going to have Y. P. Prime. It's going to be a to the eater to each of the X and Y people double prime. It's going to be a. E. To the X. Again. And so now we want to do is we want to plug this into our original equation. An original equation was Y double prime minus five Y. Prime plus six. Why equals E. To the X. And so we end up having as we have the A. E. To the x minus five, A. E. To the X plus six, A. E. To the X equals each of the X. And so that we can simplify this down. So these two terms right here, we'll make it we'll we'll translate into 18 each of the X. So have the left hand side simplifying down to A. E. To the X. And that equals to eat the X. And we can divide by each of the X on both sides. So these two terms will cancel out, so have that to A. Equals to one or that A equals to one half. And so with this knowledge, we can actually build our total solution which is the sum for homogeneous solution plus the particular solution. And so our total solution in this case is gonna be C. One E. To the negative three X plus C. Two. Eat the negative two X plus one half each of the X.

Problem or 15. Why prime minus 0.6 wives to go to zero? Why prime is you go to 0.6 Why and were given the y 005 It's OK is it goes 0.6 p zero is to go to five. Star function is wise to go to five e 20.16.

All right. So for this example, we're asked to find the solution to the second order linear differential equation. Y double prime plus four Y. Prime plus five Y equals zero. To start, we're going to find a characteristic equation which in this case is going to be or squared plus four are plus five equals zero. So next we've got to find our solution to our right. So as you can tell this can't be factored normally. So we're gonna have to use the quadratic equation. So in using the quadratic equation we're going to get R equals negative four plus or minus square root four squared which is 16 minus. And then it's going to be four A C. So four times one times five. So it's going to be 20. And then all of this over to a which is just two times one. Okay, so from here obviously 16 -20 is negative for and so whenever you take the square you can take the square root of a negative number. So really it's just going to be the square to four times I since it's negative. So essentially what we're going to have as our was negative for plus or minus to I over to Okay, which is just negative two plus or minus I. Alright, so from their negative two is are alpha. Okay, so I'm gonna write out this morning quick. Just keep in mind that's our alpha and this is beta. So now our solution is going to take the form of C one mm to the alpha two. Her sign beta two plus C two E two, the alpha T. Sign beta two. Uh So bearing that in mind our solution will take the form Y. Of T. Is equal to C one. The to the alpha two. Excuse me? C one E. To the negative to T. Because -2 is our alpha. Co sign two Plus C two E. To the negative to T. Sign T. And that is the answer to the second order linear differential equation


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