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This video from IntegralCALC shows you how to solve a sixth example of value problems.
Tags:How to Solve Initial Value Problems Example 6,integral help,integral tutor,integralCALC,integrals tutorial,math lesson,math problem equations,math tutorial,value math problem
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How to Solve Initial Value Problems Example 6
Hi everybody. One more initial value problem today, this one is dy/dx = 6e2x and the initial condition is y(0) = 10. Okay, so as always with initial value problems, the first thing we do is take the integral of the function. So, integral of 6e2x dx, and we always add dx because it's just simply part of integral notation. This little—and the dx always go together. So, it doesn’t really have anything to do with how we’re going to deal with the mathematics part of it, but we have it there.
Okay, so 6e2x. This applies every time that you have a constant in the integral because you cans ay tat this 6 and this e2x right here are multiplied together. When these are multiplied together, you can bring the 6 out in front of the integral so that you don’t have to deal with it when you take the integral. If you had e2x + 6, and that was your integral. You can't bring this out in front. You can't separate these terms because you have the plus sign operator here.
It's only when you have one term like this and these are multiplied together, can you bring it out front. So, what that’s going to look like is 6 x the integral of e2x dx and now we don’t have to deal with the 6. All we have to do is take the integral of e2x. So, let's go ahead and do that. When we’re taking the integral of anything involving e, e is kind of something with its own rules. So, these are just something that you'll have to memorize or breakdown in your chi-cheat for your test or whatever.
But, when we take the integral of e2x, normally like if we had x3, the integral of x3 would be ¼ x4. We would add 1 to the exponent and then we’d divide the coefficient by the new exponent and that would be the answer. This, since we have e2x, this exponent, you might think would be a little bit difficult to deal with when in fact all that it is and this is something that you can memorize. We have 6 out in front remember, because we’re not dealing with it. So then we just draw a big parenthesis and then we take the integral.
You only have to deal with this constant here and the x just stays. So, it's this constant 2, it's going to act in the same way that this exponent does here when we’re taking an integral of like a simple term like this.
So, in the same way that we have the ¼ out in front here, we’re going to divide this term by the coefficient on the x here. So this is going to look like ½e2x but we don’t add 1 to the coefficient here on the x term. It just stays as a 2. This never changes at all. All we need to do is divide this whole term by this number right here.
So hence, we've got this term over to 1/2, even the 2x and that’s just a rule. So, we've done that. We took the integral. And that’s all we need to do and then we add c as always because it's a constant. And this again equals y. whenever you have an initial value problem and you’ve taken the integral, you’ve solved for the function ere, you can just go ahead and say equals y.
The reason we do that is so that we have a place to plug in both of these numbers. This, we’re going to plug in zero for x and we’re going to plug in 10 for y. You have to have a place to put this 10, so you always need a y out here.
So we’re going to say 10 = and then 6 x ½ and then we’re plugging in zero for x. So e(2x0) + c. Okay, so now what we need to do is simplify. The first thing that we’ll do, we have 10 = 6 x ½ e, let's in fact, let's actually do this.
Skip one step here. We’re going to do 6 x ½. We’ll do that first, and 6 x ½ is of course 3 or 6/1 3 and then e(0) + c. Anything raised to 0 as its exponent is always 1. It doesn’t matter what it is, it could be x1, it could be the √2x + 10. It can be e0, anything to the 0 is simply 1. So, we have 10 = 3 x 1 + c which is 10 = 3 + c, so c = we’ll subtract 3 from both sides. This will cancel and we’ll get 7 over here on this side. So c = 7.
So, our final answer, let's go ahead and erase this section over here. Our final answer, we’re going to plug c. Our answer for c back into this equation right here and just simplify for our final answer.
So, I'm going to go ahead and say y = and we have 6 x ½ or 6/2, 3. So 3 e2x and then we have + c. So we’re going to go ahead and say + 7. And we’ve done all the simplification we can. We plugged in c using our initial condition. So this is our final answer.
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