Differential Equations Example 4 Hi everyone, welcome back, we’re going to be doing another differential equation of separable equation problem today. It looks like this, DY/DX=2X the square root of Y and it gives us initial condition Y of 0=9. The instructions on this problem is to actually to solve the initial value problem given here, so you may think that this is an initial value problem and it is, but the reason that it makes it a separable or a differential equations problem is because the way that we’re going to go ahead and solve it is to separate the variables on one side X and on the other side integrate and then plug the initial condition back in to come up with an explicit formula for Y. So the way that that looks, the first thing we want to do is separate the variables, so we’re going to multiply both sides by DX, we have DY=2X the square root of YxDX then we’re going to go ahead and divide both sides by square root of Y. We get DY/ square root of Y equals 2DX, so now we’ve separated the variables. We have Y from the left side; we have X’s on the right side, that’s what makes this a differentiable equations or separable equations problem is that we needed to separate the equations in order to solve it or separate the variables in order to solve it. So now that we’ve done that, we need to integrate both sides but before I do that I want to actually simplify the left side here. I need to move this square root of Y to the top of the fraction to the numerator to make it easier for me. I personally find it much easier to deal with when I have something on the numerator than in the denominator so I’m going to go ahead and convert and what that’s going to look like—of course we all remember that the square root of anything right is like the square root of X with the X to the 1/2. So this is actually Y the 1/2, lets go ahead and change that to Y and 1/2 and then of course to move it to the top all I need to do is change the signs on the exponents where right now we have a positive 1/2. If I want to move it to the top then I just have to change it to a negative, so I have Y to the negative 1/2, DY its now in the numerator as suppose to the denominator equals to DX. So I simplified to a point where I feel comfortable taking the integral, so now I take the integral on both sides. Taking me now to step two of any differentiable equations problem, first we’re separating the variables, next is taking the integrals at both sides, so take the integral to the left side, we of course add one to the exponents, so lets say Y-1/2+1 is +1/2 and then of course we divide the coefficient by which is a 1, 1 is implied there by the new exponent which is 1/2. Of course 1/1/2 is 2 so we can go ahead and simplify that right now by 2, so this is the integral of the left side, we’re done there, and we set that as equal. The integral of 2 is 2x and its x because we have x’s on this side so we have two x and then of course we always add C to the right of the equation with x’s. It’s very important that you remember to add C here at this point in the problem instead of the very end because anything—any algebra that we perform after this point needs to evolve the scene. So as soon as you integrate remember to add C to the right side. You’re only adding it to the right side, not to the left side, so go ahead and do that. Now that we’ve integrated we can go ahead and let’s simplify this. I’m just going to change why the 1/2 back to the square root of Y, so it’s a little easier to look at 2x +C, so this is our function. At this point we can go ahead and plug in the values from our initial condition because what we want to do now is plug this in so we can solve for C and then come back with an equation that is equal to Y at the very end as our answer. So let me go ahead and erase the rest of this so that—so what do we have here, we have 2x the square root of Y =2x+c, so we can go ahead and plug in our values, this of course being x and this representing