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This video from IntegralCALC shows you how to solve a fourth example of initial value Math problem.
Tags:How to Solve Initial Value Problems Example 4,integral help,integral tutor,integralCALC,integrals tutorial,math example equation,math lesson,math tutorial,value math problem
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How to Solve Initial Value Problems Example 4
Hi everybody, we’re going to keep it going with initial value problems. The next one is going to be dy/dx = 1/√x + 2 and our initial condition is y(2) = -1. So as always, with initial value problems, the first thing that we want to do is take the integral of the function that they’ve given us, so let’s do that. okay the first thing that I would like to do is move the square root of x + 2 to the top so the way that we’re going to do that is we will say, √x + 2 is the same as (x + 2)1/2. And then to move it the top, we will flip the exponent from a positive to a negative so what this is actually going to look like is the integral of (x + 2)-1/2dx. So we took it out the square root by changing the exponent to 1/2 , those two things are equal, it’s the same thing and then moved it to top by—I’m sorry this is positive now. It’s a positive ½ on the bottom and you move to the top so we change it to a negative ½.
So now that we’ve done that, we can more easily take the integral of this function so, we will say x + 2 here, we add one to the exponent, so -1/2 + 1 is positive ½ and then we divide the coefficient which is an implied one here by the new exponent which is ½. 1 divided by 1/2 is 2 so we have 2 out here. and then of course I applied the chain rule to this function to be able to take the integral in this way, the way that I’m doing it is I only deal with the outside first and dealing with the exponent and bring it out here, you know ass one to the exponent but I still have to deal with what’s inside here, I can’t just leave the x + 2 alone. I have to divide by the derivative of the inside. The derivative of this is actually only one right, if you take the derivative of x + 2, it’s 1 for the x here, the 2 goes away, it’s a constant so the derivative is zero.
So we will divide this whole function by the derivative of the inside which is 1, so it doesn’t actually have any effect on the function so we can just leave it as it is. So we have 2 × (x + 2)1/2 and then of course, whenever we take the integral, we always have to add C to back in for the constant that could potentially be there so we’ve got our function now integrated so now we need to apply the initial condition y(2) = -1, so the way that we’ll do that and remember that this whole function here is equal to y, so we plug 2in for x and -1 in for y so we will have -1 = 2 (2 + 21/2 + C) and once we plug these things in, what we’re going to try to do is solve for C. So we need to simplify so that we can solve for C.
So let’s go a head and do that, -1 = 2 times—let’s do 41/2 + C. 41/2 is the same thing as the √4. Remember we change this from the √(x + 2) to (x + 2)1/2. In the same way we can reverse it and then we have something to the ½, we can change it back to the square root. So this is actually -1 = 2 ×√ (4 + C). The √4 is 2 so we have -1 = 2 × 2 + C. so let’s go ahead and write this over here. We have -1 = 4 + C, so C = -1 – 4, we subtract 4 from both sides, it cancels over here and we have C = -5. So that’s a simple application involved in finding C. Now that we’ve found it, all we need to do is plug back in. so I’m actually going to erase this section so that I have more room to write. We’re going to plug C back in to this equation here and then just simplify a little bit s we will say, y = 2 ×(x +2)1/2 and then C is -5, so we say minus 5.
So this is our function and you could leave it this way but ½ is an exponent is a little bit messy so I would just like to change that to a square root so, I’m going to say that the final answer to this problem is y = 2 ×√x + 2 – 5. There you have it.