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Joshua C.

Calculus 2 / BC

1 year, 6 months ago

Okay, so here we're looking at, you know, f of X is equal to E to the power of our X. And we're giving the differential equation two times the second derivative of Y. It was the first derivative minus Y is equal to zero. Basically we set fx um is equal to Y. So we need is to find the derivatives first. Okay, so since E to the R x x, Y, the first derivative of that is going to be our times E to the power of our X. And then the second derivative is going to equal or squared times either the power of our X. So when I plug that in uh this equation here, so you will get to times are squared. Eat to the Rx of course our times you to the Rx minus E to the R X is equal to zero. So we can factor out the E to the power of our X. We will get to r squared plus ar minus one is equal to zero. Yes, we can factor that a little bit further, so too are square plus ar minus one. More factor too two. Ar minus one times our plus one mm. Since that is equal to zero and you know the Rx certainly is going to equal zero. Um we get that are one is equal to one half and our two is equal to negative one. Yes, this is the incident for a part A. And then we go to part B, which tells us that we can take the ours we found and make a new Y. In this case, why is going to be equal to A times eat our first ourselves community. X over two plus B times E. Um to the power of our second are negative one times excellent negative X. We need to show that this function is, it is also solving differential equations. We're going to do that by showing again that with this. Why are too second derivative of the second derivative plus the first rate of minus Y is equal to zero. So that's what we have to show. Um So to get there we need to take some derivatives. The first decade of this is going to equal one half times a times E. To the power of X over two minus B times E. To the negative X. And the second derivative is going to equal one force times a times B to the power of X over two plus B times a year. The power of negative backs. And I just have to add these up in this way. Okay, so if we want to times the second derivative, I think this one half times a times E. To the power of X over two plus two times B times eat and negative X. And we want to add that to the first derivatives. We're going to add one half times a times E uh X over two minus B times E. To the power of negative X. And then we need to subtract why from this. Okay. So subtracting why we get negative A times E X over two minus B times eternity. The X spending some cancellations. Right, So here we have a positive one half, uh huh. A positive one half A and a negative one. So that's going to equal zero. We also have for the bees to minus one minus one, it's going to equal to zero. So the whole thing equals zero. That's what we wanted to show. We needed to show that in order for this to be a solution that two times the second derivative plus the first derivative plus Y equals zero. So that checks out and that finishes finishes the problem. We have now shown that it is indeed a solution

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Lectures

01:11

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is, loosely speaking, a formula that describes the function. Integration is the inverse operation to differentiation. The integration of a function "f"("x") is its indefinite integral denoted by ?"f"("x")dx.

06:55

In grammar, determiners are a class of words that are used in front of nouns to express how specific or non-specific the noun is. A specific noun is one that can be identified in a unique way from other things. For example, "the cat" is a specific noun, while "a cat" is an indefinite noun, because it is not clear if it is the same cat that was already mentioned, or another cat.

01:14

For what values of $r$ does the function $y=e^{r x}$ satisfy the differential equation $y^{\prime \prime}-4 y^{\prime}+y=0 ?$

03:42

(a) Show that every member of the family of functions $ y = $ (In $ x + C)/x $ is a solution of the differential equation $ x^{2} y^{'} + …

02:02

For what values of $ r $ does the function $ y = e^{rx} $ satisfy the differential equation $ y" - 4y' + y = 0? $

01:56

(a) Show that every member of the family of functions $y=(\ln x+C) / x$ is a solution of the differentialequation $x^{2} y^{\prime}+x y=1$…

02:51

A student takes a 32-question multiple-choice exam, but did not study and r…

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