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Integrate initials order

in this video we're going to talk about how to solve the initial-value problem as it relates to differential equations so let's start with this example problem first let me adjust the size on this so let's say that dy over DX is equal to 6x minus 3 and we're given the point y of 0 is equal to 4 so this means that x is 0 and Y is 4 how can we solve this differential equation the first thing I'm going to do is multiply both sides by DX so I have dy is equal to 6x minus 3 times DX so I'd like to separate the variables I like to separate Y from X once the variables are separated we can integrate both sides so the integral of dy is simply Y the antiderivative of x or X to the first power is x squared over 2 times the constant that was in front of it which is 6 the antiderivative of a constant is just that constant times X and of course we need our constant of integration C so this right here is the general solution to the differential equation but now we need to plug in our initial value and the reason why we're going to do that is so that we can find the value of C so we said that Y is 4 6 divided by 2 is 3 X is 0 and so 0 squared times 3 is 0 so all of this is 0 we can clearly see that C is equal to 4 now the last thing we need to do is plug in the value of C so our final answer is going to be 3x squared minus 3x plus 4 so this is the solution to the initial value problem now let's move on to our next example so let's say that dy over DX is equal to 2xy and let's use the initial value Y of zero is equal to let's say to go ahead and work on that actually let's make that three now once again let's separate the variables but first let's multiply both sides by DX so we have dy is equal to 2xy times DX now the next thing that I'm gonna do is the viable sides by Y so on the Left I have 1 over Y dy on the right 2x DX so all of the variables have been separated we have the Y variables on the left side the X variables on the right side now the next thing that we need to do is find the antiderivative of both expressions the antiderivative of 1 over Y is the natural log of Y that's one of those things you need to know the antiderivative of X is going to be x squared over 2 and then plus the constant of integration so canceling the two we have Ln Y is equal to x squared plus C now what do you think we need to do at this point what we need to do is place both sides of this equation on the exponent of e so we're gonna have a race to the ln y is equal to e raised to the x squared plus C now let's talk about why we can do this e is equal to e so Ellen y must be equal to x squared plus C which is what we have here so it works we can do that now it turns out that e raised to the Ln Y the natural the base of the natural log function is e this simplifies to Y now how can we simplify that expression review an algebra we know that x squared times X to the third is X to the two plus three which is X to the fifth power so working in Reverse let's say we have X raised to the 4 plus five we can write that as X to the fourth times X to the fifth so e x squared plus C can be written as e to the x squared times e to the C but I'm going to write this one first so we have e to the C times e to the X word e is a constant C is a constant a constant race to a constant is a constant so we can simplify our answer like that just so you know this C is different from that scene they're not the same so you can think of this as lowercase e and maybe that as uppercase e now let's plug in our initial value y is equal to three X is equal to zero anything raised to the zero power is one so therefore C is equal to 3 in this example so the solution to the initial value problem is 3 e raised to the x squared and so that's it so now you know how to solve the initial-value problem as it relates to differential equations