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Answers

$$ \begin{array}{l} \text {
} \frac{9}{z^{2}+2 z-8}, \frac{4 z}{z^{2}-4} \\ \text { LCD }(z-2)(z+4)(z+2) \end{array} $$

So I'm gonna just combine them because I do have the same denominator, so I don't have to do anything else it will be. Don't be a squared minus three W plus six plus nine minus w squared which would becomes of these w squares. Cancel out. And then these combined to 15. So got negative three w plus 15 over W minus five. And then from here, I'm gonna factor out of negative three from the top of modern to get W minus five. You can see how the top bottom cancels out. And the answer to my 1st 1 should actually just b negative three. Same with the 2nd 1 I can already just combine because the top in bar that bottoms are the same. So then I would do choosy. Squared minus C square disease squared. Negative three Z minus negative. Five z gives me to Z six months. Nine gives me negative three. Oliver C. Squared minus one. I'm here. I'm gonna factor both the top and bottom. The bottom is easy. That's just a difference of square. So it's C minus one seat plus one. Um, the top. It looks like it's gonna be easy um, plus three, the minus one. So then the Z minus one on the top and bottom will cancel out, leaving me with C plus three or Z post with.

And so here we have the equation Third derivative of the U minus two times thesis, second derivative plus the first dreaded is equal to X minus e to the power of X. So the corresponding homogeneous equation as the associated auxiliary equation of our cubed minus two R squared plus r. So we want a factor that we'll get r minus one Squared times are that means our is equal to zero and one with a multiplicity of to or rather, two occurrences. So our general homogeneous solution and that being some constant times here, the power of X plus some other constant times x times e the power of X plus a constant kiss. And I'm gonna let's f of x be the right hand side x minus E two x on and we get then that the first derivative is equal to one minus e to the X. He second, derivatives is going to be equal to negative Eat of the X on the third derivative is equal to negative year the x so we can take the third derivative Subtract The second driven did not give us zero. Some things are operator is gonna be the cute minus D squared on F ticket zero Annihilator It's gonna apply De que dynasty squared on to the sea The third river of Z minus two times the second derivatives E plus the first derivative. And we end up with the Sixth Street Live minus three times the fifth derivative plus three times the fourth derivative minus the third derivative. Um, we apply this to the right hand side. It becomes zero that since we chose something that would do that intentionally Okay, so now this associate ID on Zilly area equation is our how to six times three r to the power of five plus three R to the power of four, plus our cube. So we factor that and we get r minus one cubes times are cubed. So we have r is equal to zero with multiplicity three and is equal toe one with multiple city three. So this means our general solution is equal to see one times either the x plus c two x e to the power of rex plus c three x squared times e to the power of X plus C for plus C five x plus c six x squared roll the Caesar Constance. So then, from above are homogeneous. Solution was see one either the X plus c two x e x plus C three. So that gets the particular portion We're going, Teoh, take the full thing and so attract the homogeneous part. So we see our homogeneous portion have a C one time constant times e to the X, and we have that here. They had a constant times x times E to the X. We had just a constant. It's just gonna keep everything that's not underlined and red then and we have C three times x squared E to the X plus C five times X plus C six times x squared, and that's gonna be the particular solution.

In this video, we're gonna go through the Anson's question too benign. Chapter seven, part five to rest. If all of us to solve this initial value problem, that's a first, uh, so we want to use initiative to use look, less transforms. But it's only really gonna work if we have initial values, huh? To use equal zero. Okay, so less change the time. Very boob. So let how equal to t minus what? The towers under your time variable. It's just a, uh mm. Subtracting one from the time. So the derivatives with respect t are just gonna become derivatives with Spends time without changing anything. Um, so now, in terms of towel weaken right, this equation as, uh, everything is the same. Except we've just got tower here instead of t minus one on DDE. Uh, initial conditions in our violated zero rob someone. Okay, so now we can use Plus transforms, like, usual s o. We know what the the chancellor's insect intuited. Uh, so if big set capsules, that is the bus transport little that and it's minus minus. Want on this minus nine? Because five times questions. All of it is that crime, which is s times said, Plus well, minus six Zed. Well, big said then. 21. Last chance for 21 each. Tower 21. Over. Yes, Money. Okay, that's a no rearranging this four bigs that we've got 21. I guess that's what that affect all the, uh, terms and left inside that? No. Ah, it says a big said So it's gonna be s minus nine course. Fine. They went to divide by, um, over coefficients off the sets that said s squared was five s minus six. Okay, now let's try it. Sort this out. Let's most bottom bottom by X minus one. We're gonna get 21 minus that spineless fool. That's minus one on the denominator guys minus one. And then this X squared plus five minus six. Come fact arised the s cost six asked, minus one. So I made that square. Okay, so now we can use past affections again. I'm not gonna go through the details for this, but it's just that's standard procedure. Is that piquancy? Three of s minus one square, minus one of escorts. Six. And we know what the what? The investor class transforms off these. That's gonna tell us that Zed, as function of Tao is three e to the t times T minus eight months. Six t. Okay. Uh oh, hang on a minute. Thes tease. Thes t should have been toes, because now we can change back to the original variable T. So is that its function of tea is now three into the tea, minus one. Ties by T minus one minus mmm to the minus six. So you might as well. And that's the solution to this differential equation.

In this problem, we're being asked to identify any pairs of like terms. Well, let's start by talking about what late terms our life terms have to be two terms to have the same variable along with the same experiment. So for example, nine A and A squared would not be considered like terms because a square has an explanation too. And for this 98 term, the A's exponent is one. So because they are not the same exponent, those are not like terms. So let's see if we can identify any pairs of lake terms. Well nine a doesn't have a late term because no upper term has just in a for variable a square doesn't have a late term because no other term has a squared. Now let's look at 16, it doesn't have a variable. We'll discover term four doesn't have a variable as well. So those who are considered like terms so 16 and four next 16 B squared. Well, it's like term would be nine B squared because they hope that the variable B squared, so 16 B squared and nine B squared. And now we've identified all are pairs of like terms. Okay,


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