Solve the initial value problems in Exercises 53–56 for y as a...

Question

Solve the initial value problems in Exercises 53–56 for y as a function of x.

√(x² - 9) (dy/dx) = 1, where x > 3, y(5) = ln 3

Accepted Answer

Hello there, today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Solve the initial value problem. The square root of x 2 minus 1 multiplied by Dy by dx is equal to 1, where x is greater than 1. Y 2 is equal to 0. OK, so it appears for this particular problem, we're asked to solve for the initial value problem using the information that is provided to us for the prom itself. So now that we know that we're solving for this initial value problem. Our first step that we need to take is we need to begin by separating the variables. So, with that said, we need to begin by isolating DY by DX, and we need to separate our variables. So we need to take the square root of x 2 minus 1, multiplied by DY by DX is equal to 1. And we need to divide both sides by the square root of x 2 minus 1, and when we do, we will find that it's gonna be equal to DY divided by DX or DY by DX is equal to 1 divided by the square root of x to the 2 minus 1. Fantastic. So now, we need to separate the variables by moving DX to the other side, so we can therefore write that DY is equal to 1 divided by the square root of x 2 minus 1 multiplied by DX. Our second step now is we need to integrate both sides. So we now need to integrate both sides of our equation. So we need to take the integral of DY is equal to the integral of 1 divided by the square root of X2 minus 1 multiplied by DX. Fantastic. And we need to note that the left-hand side of our equation will be a straightforward answer. We will simply get that the left-hand side is equal to Y. We take the integral of DY, so Y will be equal to the integral of 1 divided by the square root of x 2 minus 1. Multiplied by DX. And the integral on the right-hand side, as we should be able to recall and recognize, is a standard integral, which we can evaluate it as follows. So we could take the integral of 1 divided by the square root of x 2 minus 1 multiplied by dx, which will be equal to the natural log of the absolute value of. X plus the square root of x 2 minus 1. Plus C where C is some constant value. So therefore, with that in mind, our solution to the integral will become y is equal to the natural log of the absolute value of x plus the square root of x 2 minus 1. Plus C. fantastic. So, our third step, we're on the step 3 now. Our third step is we need to recall and apply the initial condition. So we are told by the prom itself that the initial condition is Y of 2 is equal to 0. And when we substitute x equal to 2 and y of 2 is equal to 0, back into our equation, we will find that we're able to write, and once again we're plugging in a value of. So once again we're told that Y of 2 is equal to 0, so when we plug in a value. Of 0 in for Y. And a plug in a value of 2 in for x, we will be able to write that it will be equal to 0s equal to the natural log of the absolute value of 2 plus the square root of 22 minus 1. Plus C. Fantastic, and when we simplify. Or begin to simplify, we will find that 0 is equal to the natural log of the absolute value of 2 plus the square root of 4 minus 1 + C. Which, when we simplify some more, we will find that it'll be, we could write that 0 is equal to the natural log of the absolute value of 2 plus the square root of 3. Plus C, fantastic. So now we need to rearrange this expression to isolate and solve for C, and when we do, we will find that C is equal to negative, the natural log of the absolute value of 2 plus the square root of 3. OK. So now we need to substitute C back into our general solution. So we need to note that Y will be equal to the natural log of the absolute value of x plus the square root of x 2 minus 1 minus the natural log of the absolute value of 2 plus the square root of 3. And when we simplify, we will find that Y is equal to the natural log of parentheses of the absolute value of X plus the square root of x 2 minus 1. All divided by 2 plus the square root of 3. And we need to note that since X is greater than 1, and we need to note that X plus the square root of X2 minus 1 is greater than 0, or X is greater than 1, that means we can safely drop the absolute value signs, so we can get rid of the absolute value signs. And with that in mind, we can go ahead and write that y is equal to the natural log of parentheses of x plus the square root of x 2 minus 1, all divided by 2 plus the square root of 3. And this is our correct and final answer. Hooray, we did it. So we solve for our final answer. So going back to the top to look at the prom itself and to recap what we've learned. We could safely say that our final answer for the initial value problem once again is that y is equal to the natural log of parentheses of x plus the square root of x 2 minus 1 all divided by 2 plus the square root of 3. And that's it. Thank you so much for watching, hopefully that helped, and I can't wait to see you in the next video. bye.

Solve the initial value problems in Exercises 53–56 for y as a function of x.√(x² - 9) (dy/dx) = 1, where x > 3, y(5) = ln 3

Watch next

Solve the initial value problems in Exercises 53–56 for y as a function of x.
√(x² - 9) (dy/dx) = 1, where x > 3, y(5) = ln 3