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 + 1)² (dy/dx) = √(x² + 1), where y(0) = 1

Accepted Answer

Hello there, today we are going to solve for 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. Parentheses of X2 + 42 multiplied by DY by DX is equal to the square root of x 2 + 4. Y 0 is equal to 2. OK. So it appears for this particular problem, we're ultimately asked to solve for the initial value problem that is provided to us by the problem itself. So, using all the information that is provided to us, we're asked to solve for the initial value problem. So now that we know what we're ultimately trying to solve for, our first step that we need to take is we need to separate the variables. So, as we should recall, we need to begin by separating the variables in our equation. So, we need to take parentheses of X2 + 42 multiplied by DY by DX. And this will be equal to the square root of x 2 + 4. Now, we need to divide both sides of the equation by, and let's make a note, we need to divide both sides of the equation by parentheses x 2 + 4 in parentheses to the power of 2, and we need to multiply both sides by D by DX. So we are going to multiply, which I'll use a little dot symbol to remind us, we need to multiply both sides by DX2. And when we do that, we'll be able to write that DY5. DX is equal to the square root of X2 + 4, divided by parentheses X 2 + 4 in parentheses the power of 2. Fantastic So now, at this stage, we can separate the variables by bringing DY to one side and the terms involving x to the other side. So we could go ahead and write DY is equal to the square root of x to the 2. + 4 divided by parentheses of x to the 2 + 4. To the power of 2 DX. Fantastic. And when we simplify by recalling and using the right, so once again, once we simplify using our techniques that we should be able to recall, we could simplify the right-hand side to write that DY is equal to 1 divided by parentheses of x 2. Plus 4 to the power of 3 halves, or 3 divided by 2 multiplied by DX. Fantastic. Our second step now, step 2, so moving right along here, step 2 is we need to integrate both sides. So now we need to integrate both sides of our equation, and when we do, we will take the integral of Dy, which will be equal to the integral of 1 divided by parentheses x to the power of 2. Plus 4 to the power of 3 divided by 2. DX And we need to note and recall based on our prior knowledge that the left hand side is a straightforward. Equations will result in a straightforward equation because we will find that y will be equal to the integral of 1 divided by parentheses of x 2 + 4 to the power of 3 divided by 2. DX Our 3rd step now, this is step 3, is we need to solve for the integral, and in order to solve for the integral, we will need to recall and use a standard result for integrals in the form of. And this will be the form you need to recall how to use. The integral of 1 divided by parentheses, X2 + A 2. All, so once again, in parentheses, x 2 plus a 2, will all be to the power of 3 divided by 2 DX, which will give us drum roll. When we apply this form, we can go ahead and write the integral of 1 divided by parentheses of x 2 plus a 2, all to the power of 3 divided by 2, so. 3 divided by 2 multiplied by DX, which will be equal to drum roll. X divided by a 2 multiplied by the square root of x 2 plus a 2. Plus C. And we need to note that in our case, let's make a note in blue, in our case, A will be equal to 2. So because A is equal to 2, the integral will become as follows, the integral of 1 divided by parentheses of x to the 2 + 4, all to the power of 3 divided by 2. Multiplied by DX will be equal to X divided by 4 multiplied by the square root of X2 + 4 + C. Fantastic. So, moving right along here, thus we could go ahead and write that the solution to the integral is y is equal to x divided by 4 multiplied by the square root of x 2 + 4. Plus C, fantastic. So step number 4. Is we need to recall and apply our initial condition. So, we need to note that we are given that Y is equal to, so specifically, we need to note that the problem itself states that Y of 0 is equal to 2, meaning that this function of Y of X, where X in this case, or our variable is 0, so Y of 0 is equal to 2. And when we substitute X is equal to 0. And we substitute in that y of 0 is equal to 2 into our equation, we will be able to go ahead and write that 2 is equal to 0 divided by 4 multiplied by the square root of 0 to 2 + 4. Plus c, and once again we're plugging in a value of zero in for x, and when we do, we will find that 2 is equal to c. So that means that without a doubt the constant of integration will be c is equal to 2. And once again this is our constant of integration. So with that said, Moving right along here, we now need to substitute C is equal to 2 into our general solution, and we will achieve this particular solution. We will achieve that y will be equal to x divided by or multiplied by the square root of x to the 2 + 4. Plus 2 and this will be our correct and final answer. Hooray, we did it. So going back to look at the prom itself to recap what we've learned, we can safely say that our final answer for the initial value problem is once again. That Y is equal to X divided by 4 multiplied by the square root of X2 + 4 + 2, and this is our correct and final answer. 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 + 1)² (dy/dx) = √(x² + 1), where y(0) = 1

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Solve the initial value problems in Exercises 53–56 for y as a function of x.
(x² + 1)² (dy/dx) = √(x² + 1), where y(0) = 1