In Exercises 1–4, show that each function y=f(x) is a solution...

Question

In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.

3. y = 1/x ∫(from 1 to x) e^t/t dt, x²y' + xy = e^x

Accepted Answer

Welcome back everyone. In this problem we want to find the first order differential equation of the form x2y plus 2xy equals g x, where y is the derivative of y, for which the function ye equals 1 divided by x2 multiplied by the definite integral between 1 and x of the cosine of t with respect to t is a solution. Now we're trying to find the first order differential equation in this form, and if we think about it, we already know what y is. So if we can figure out the derivative of y, OK, then we should be able to substitute it into the left hand side of the differential equation to figure out our solution. So let's go ahead and see if we can do that. So far we know that why. is equal to 1 divided by x2 multiplied by the integral between 1 and x of the cosine of t with respect to t. So why is a product, product of two expressions. That means to differentiate, we can use the product rule of differentiation. Recall that by the product rule, it says that if we have a function y that's equal to u multiplied by v, where each term u and v are functions of x. Then the derivative of y would be equal to. UV plus UV. In other words, we differentiate both terms and multiply them by their opposite original term, by the opposite original terms. OK, so for our problem here, if we let u equals 1 divided by x2, we can differentiate to figure out the derivative of u, OK. And here if we rewrite this as x -2, then the derivative of u would be equal to 2 x 3, or we can rewrite that as 2 divided by x cubed. On the other hand, if we let Z. Equal the integral between 1 and x of the cosine of t with respect to t. OK, then if we integrate it, sorry, we differentiate it, the derivative of f. Here is going to be equal to the cosine of x. And let me just fix this sign here because we're saying earlier the derivative of you you prime. So now that we have U and V, we can apply the product rule, which means that Y will be equal to -2 divided by x cubed multiplied by the integral between 1 and x of the cosine of t with respect to t, that is uuV plus UV, that is 1 divided by x2 multiplied by the cosine of x. Now we can substitute this expression back into our the form of our first order differential equation. So this means then. In the left hand side of the differential equation, it will be x2 multiplied by the derivative of y. Yeah, which we can rewrite here. OK. Plus 2 x y and again we know that Y is 1 divided by x 2. Multiplied by or integral. Now, before we equate it to G of X, let's try to simplify our expression a bit, OK. Now, to do that, we can distribute our terms. And we'll end up with 2 x 2 divided by x cubed multiplied by the integral. OK. Plus x2 + x divided by x2, OK, plus 2 x divided by x2 multiplied by the integral. Or rather or integral that we've been uh referring to since the beginning of our problem. Now notice we have the opportunity to cancel some terms here. So when we do that, OK, we're left with -2 divided by x multiplied by the integral between 1 and x of a cosine of t with respect to t. Here are x2 terms cancer to leave us with a cosine of x, and here again x cancels from x2 to leave us with 2 divided by x multiplied by our integra. Now if you notice here, our integral terms cancel out perfectly. Because this is the negative version and this is the positive version of the expression, so we are simply left with the cosine of x. Therefore, g of x is equal to the cosine of x, and that is. If we were to rewrite this, that means then that X2 Y, OK. Plus 2 xy is equal to the cosine of x. This is our. First order differential equation of this form for which the function Y is a solution. Thanks a lot for watching everyone. I hope this video helped.

In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.3. y = 1/x ∫(from 1 to x) e^t/t dt, x²y' + xy = e^x

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In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.
3. y = 1/x ∫(from 1 to x) e^t/t dt, x²y' + xy = e^x