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2 (a) Consider the third order linear homogeneous differential equation dy dy drs + 3d22 + 3d2 +y = 0. Write down the characteristic equation for (1). Factorize the...

Question

2 (a) Consider the third order linear homogeneous differential equation dy dy drs + 3d22 + 3d2 +y = 0. Write down the characteristic equation for (1). Factorize the third degree poly- nomial in this equation; you may use the MAPLE command factor() here, if you like. Using the relevant material from class and explaining briefly the procedure; write down the general solution of the DE (1).

2 (a) Consider the third order linear homogeneous differential equation dy dy drs + 3d22 + 3d2 +y = 0. Write down the characteristic equation for (1). Factorize the third degree poly- nomial in this equation; you may use the MAPLE command factor() here, if you like. Using the relevant material from class and explaining briefly the procedure; write down the general solution of the DE (1).



Answers

Second-Order Potpourri For each of the following second-order differential equations, find at least one particular solution. You will need to call on past experience with functions you have differentiated. For a significantly greater challenge, find the general solution (which will involve two unknown constants)
(a)$y^{\prime \prime}=x$
(b)$y^{\prime \prime}=-x$
(c)$y^{\prime \prime}=-\sin x$
(d)$y^{n}=y$

(e)$y^{\prime \prime}=-y$

Yeah. So for a question area have divide over the X dressy calls to the integrate off 12 X plus four. So d y over the X just equals 26 Squire plus four x with a constant. So integrate both sides. We'll have y equals two to act to the power of three plus two x choir. Let's see one x You forgot the X here, Reese constantly to and for B We'll have integrate off the term Amanda life that that's just equals to the y over the X So we can just write you I over the x here integrate offi to paragraph X sign next the X so D y over the X just equals two. With our constant and now less integrate, both sides will have white just equals two c one x with C two. Yeah, and foresee We'll have the y over the X equals to integrate off extra power of three actual power off minus three. So do you buy over The X just equals two x two powerful over full minus 1/2 extra power to plus the constant. And we integrate both sides. Okay, we can see that it's the integrate off the y. And this is the integrate. Yeah, See, Wan T X. So why Jesse kills Thio acts the power off 5/20 Last one hour two x and constancy.

So for a we have the relative off y equals two X So why just see calls to integrate equals two x choir over two plus constant And for B we have a drought Thio off y equals two minus x So you have the y over the ACSI calls to minus X and then we integrate Both sides will have integrate off the y e house to integrate off minus x the X So why just ecause too minus X choir over two plus the constant. And for see, we have the relative off. Why equals two? Why So do you buy our d? Xy goes to why we'll have the integrate off one over y the y ico sued integrate off the X So we'll have y equals to see times e to the power off X and 40 Here we have white prom Mikhail Su minus Why, the same as the last one will have y equals to see times e to the power off minus X And for e we have Yeah, duality off. Why? He tells you x y So the integrate of one over Y d y just ecos to integrate off X t X so long, Why Eco's thio x The power off to over to and we'll have white plus a constant. So why here? Jesse goes to see times e to the power off x choir over to

Okay, so this question we're asked to find the original differential equation. And so to do that, let's go ahead and rewrite this A C one E. To the zero X plus c two E two the two X. And in doing this, we're going to be able to identify our values easier. And so let's go ahead and do that. We'll have our equals two, zero and two. So this means are factors were R times R -2 And it's equal to zero. So we can expand this a little bit and doing so well. Have r squared minus two are equals to zero and we can go ahead and replace these are scored in our values with our original substitution, which would have been y double prime -2, y prime equals to zero. So for this question this is going to be original differential equation and or an answer.

Okay. So for this question we have to find the original differential equation. And so start off, let's go ahead and identify our values And our values in this case are -4 and -3. We can find them by looking at this power right here. And so from this we can go ahead and identify what original factors were And that was our plus four times arm plus three And it equal to zero. So we can go ahead and expand this. Who have R squared plus three, R plus four are Plus 12 equals to zero, or R squared plus seven, R plus 12 equals to zero. And we can go ahead and replace the czars with Wise. So we'll have Y double prime Plus seven y prime plus 12, 12 Y equals to zero. And this is in fact our original differential equation.


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