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Verify that the given function is a solution to the given differential equation $\left(c_{1} \text { and } c_{2}\right.$ are arbitrary constants), and state the maximum interval over which the solution is valid.$$y(x)=c_{1} x^{2} \cos (3 \ln x), \quad x^{2} y^{\prime \prime}-3 x y^{\prime}+13 y=0$$.

is the solution

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 2

Basic Ideas and Terminology

Differential Equations

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in this problem were given a differential equation over here and a possible solution. Why FX here? And we want to see if this is a solution to this differential equation. So first, to do that, we need to make sure that wise at least twice differential, which it is, is a function of multiplying power of X with co sign and natural log of axes, air all differential functions so we can go ahead and start taking the derivatives. So we will start with White Prime of X. We're taking the derivative of this, um, expression right here. And so we're gonna have to do the product role and the chain role here. So this is going to be to see one X times. We'll leave the second term alone. Co sign three natural log of X and add it. Teoh, First term left alone times the derivative of the second term is gonna be negative. Sign three times the natural log of X times, the derivative of three. Ellen of X is three times one over X. All right, we can go ahead and simplify the second term and to move this three up to the front so it's moved that right here. That was sloppy. Um, and one over X and X squared are going Teoh Cancel to be just X. So this is what we get Now we can take the second derivative. Why? Double Prime of X is equal Teoh again. We're gonna have to do the product rule in the general, um, to see one Times co sign three natural log of X plus, uh, two c one x negative sign three natural log of X times three times, one over X and now we need to take the derivative of the second part. So that is going to be minus three C. One times, sign three natural log of X. And now we take the derivative of the second part of this expression while leaving the first word The same. So plus three c one x sexually going to be a minus, not a plus because of that negative sign. And there so minus three C one x co sign three natural log of X times three terms. One over X, the derivative of this inside natural log here. All right, um, we'll go ahead and rewrite that a little bit more simplified and get to see one co sign of three natural log of X plus Teoh. I'm sorry. Plus, there's six in here and there's actually a minus sign. So let's make this minus six. See one. The one over the X one over X and X will cancel and become one. So we just get times sign three. Natural log of X in this next term. Nothing to simplify. Minus three C one sign three Natural log of X and finally simplifying This term we get, um, negative. Three times three is negative. Nine one over X and X cancel become one. So native nine c one Times Co sign three Natural log of X. All right, So now we have simplified expressions for why double prime white prime and for why of X, and we can plug them into our differential equation here and make sure we get zero. So let's go ahead and do that. Will do that in blue, starting with X squared times y double prime or expression for why double prime is going to be to see one co sign of three natural log of X minus six C one sign three natural log of X. This is the parentheses, um, minus three. See? One sign three Central log of X minus nine. See one co sign three natural log books. Okay. And then we want to subtract three x times widen prime. So why, Prime is to see one x Times Co. Sign three natural log of X out whips plus well, actually, will be minus nine C one x times sine of three. Natural log of X. And finally, we want to add 13 wide to this. So plus 13 or expression for why is see one X squared co sign with three times the natural log of X. All right, now, we want to make sure the simplifies and that we get zero when it does. So let's go ahead and distribute out everything we have. So here we have to see one X squared Times Co sign three natural log of X minus six C one X squared Sign of three times the natural log of X minus three C one X squared Sign times three turns The natural log of X. Okay, Um, I'm just double checking that That seems right. And yep, it seems like we can go ahead and combine. Something's actually, um but we can we'll just do that later. Okay? Um and then we have minus nine C one x squared co sign three natural log of X. Okay. And then distributing this next one, we get minus six C one X squared times co sign three natural log of X minus nine C one X squared Sign of three times the natural log X and we, lastly, are just going Teoh, put that 13 on the C one X squared co sign three natural log of X. So now we have two types of terms. Here we have, um X squared Times co sign of three natural log of X and X squared sine of three times the natural log of X. And so we just kind of need to count them up. So here we have negative six C one X squared time. Sign three natural log of X. Here. We have minus three of them. Um, and here's minus nine more. And here Oh, that's a cousin. So here, uh, we should have a This should be positive because, uh, I for we forgot to distribute this negative sign this negative three x so that's my fault. This should be plus in blue will make this plus here. Okay, so and then we get this minus six in this minus three of them. Cancel with the positive night of them. That's good. All of these terms go away. Now we count up the coast signs. So we have negative nine and negative six, which is negative, 15 of them. And then we have 13 plus two of them, which is positive. 15 of them. So we have negative 15 plus positives. 15 C one times X squared temps co sign of three times the National of X. So all of these cancel and we get this expression is equal to zero, which is good. That means that this Y of X is a solution to our differential equation. The last thing we need to do is say for what X, That is true. Well, all of the exes and are Why are y prime and are wide, double prime are going into functions of X squared and co sign sign and natural log of X. The only one of those that has certain stipulations on what exes can go into the function is the natural log of X, and in that case, our ex needs to be positive. So Rx can't be anything negative or zero. Um, but that's going to be our only stipulation. So why of X is a solution to this differential equation? As long as X is positive or we can write X is in between zero and infinity in the real numbers.

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