 We are now going to talk about integration using initial conditions and particular solutions. If you take a look at the graph here, you'll notice these are all the same shape. What we have here are the general solutions of an antiderivative. The antiderivative is x cubed minus x minus 2. And if you take a look at these graphs, they all obviously are cubic graphs. Notice if you look closely, though, all of these curves, although they have the same shape, pass through different points on the y-axis. That's because the constant of integration, the c-value, would be different for each one of these curves. The variable part is identical for all of them, but what differs is the constant, which of course we know, affects the height of the curve in the coordinate plane. You'll notice one of these curves running right here through the point 0 negative 2 is in bold face. That is the case in which we were given an initial condition, specifically that the point 2 comma 4 had to lie on the original curve. You can see it up here on the upper right, the point 2 comma 4 right there. Anytime you're given an initial condition, a condition that has to be true of the original function, we are able to find a particular solution to the antiderivative, meaning we are actually able to find what that plus c-value is. And we're going to take a look at examples of how you do this. In this first one, we are asked to find the general solution of f prime of x equals e to the x and then find the particular solution given that capital F of 0 is equal to 3. Notice we are first asked to find a general solution. That implies we want an answer that has plus c in it. In order to find the original function, we need to take the antiderivative of e to the x. So the original function f of x is going to be found by doing the antiderivative of e to the x dx. Notice the notation I am using to set this up. So think of the rules we know. Antiderivative of e to the x is simply e to the x, but don't forget your constant of integration. This is where we can now use the condition we're given, the initial condition, that f of 0 had to equal 3. Well 0 is your x, we can simply put that in for x and e to the x plus c. f of x has to equal 3, the function needs to equal 3. You are essentially just setting up a little equation to allow you to solve for c. e to the 0 of course is 1, so c is equal to 2. So what that means in the end is that our function f of x is e to the x plus 2. This is what we refer to as a particular solution. We have solved for our constant of integration and we know its exact numerical value based upon this initial condition. Let's look at another one. In this problem at any point x comma y on a particular curve, the tangent line has a slope of 4x minus 5. If the curve needs to contain the point 3 comma 7, find its equation. This is a great problem because of how it's worded. Notice you are told the tangent line has a slope of 4x minus 5. That's your derivative. By definition, derivative is slope of a line tangent to a curve. You're being given what the derivative is. So if I want to find the original curve, my original function f of x is simply going to be found by taking the antiderivative of that derivative. Again, notice how I'm setting this up. This is considered the correct and proper mathematical notation for setting up a problem such as this. Antiderivative of 4x we're going to keep the 4 multiplied by x squared over 2 minus 5x plus c. Now you know the point 3 comma 7 is on the curve. Let me go ahead and simplify this a little bit before we move on. Obviously quadratic. We are told the point 3 comma 7 needs to lie on this curve. So the function value is 7. If we substitute 3 in for our x's, this gives us 18 minus 15 plus c. We end up that c is equal to 4. So what that means is that our function is 2x squared minus 5x plus 4. Again, this is what we refer to as a particular solution based upon the initial condition we were given that the point 3 comma 7 had to lie on the original curve. Here we are asked to find the equation for y. Given that dy dx is equal to negative 1 over x squared and also the fact that the point 1 comma 3 needs to lie on the curve. What you are given there in dy dx equals negative 1 over x squared that is a typical differential equation. Any kind of equation that's given to you as a derivative in this case dy dx equals negative 1 over x squared that is a differential equation. In order to find y we are going to take the antiderivative of this derivative using the fact we know that antiderivatives and derivatives undo each other therefore we will arrive at the original curve. So in order to find the function y we need to do the antiderivative of negative 1 over x squared dx. We can rewrite this in order to do our antiderivative as negative 1 x to the negative 2. The negative 1 is going to stay as our coefficient. We need to add 1 to the exponent and divide by it. So when we do that we end up with x to the negative 1 then we are dividing by negative 1. So negative times negative is positive. You can leave it as x to the negative 1 or if you care to rewrite it as 1 over x that is fine as well. So that is our general solution for our equation 1 over x plus c. But now we can use the fact that we know the point 1 comma 3 lies on the curve. 3 is the y value, 1 is the x. We obviously get a c value a constant of 2 so that means our function in the end is 1 over x plus 2. Again this is what we refer to as the particular solution for the initial condition that the point 1 comma 3 had to lie on the curve. This last example is perhaps the most complicated of all of them. We are asked to solve the differential equation. The second derivative is given to be x plus squared of x given that f of 1 is 1 and f prime of 1 is 2. So in this case we're given a second derivative. We have to work our way all the way back to the original function from the second derivative. So as you can imagine this is going to be a multi-step problem because if we take the anti-derivative once we arrive back to our first derivative. If we take the anti-derivative again that's how we arrive back to our original function. So this is going to have multiple steps to it. So let me try to number them for you just so we can keep ourselves organized. So the first thing we're going to do working from our second derivative we're going to take the anti-derivative of that because that's how we can arrive at the first derivative. And I'm going to go ahead and rewrite that square root of x as x to the 1 half. Adding 1 to the exponent we get x squared over 2. Adding 1 to the exponent of 1 half we get 3 halves but then we have to divide by it so that's the same as multiplying by 2 thirds. So we have here a general solution because we have a plus c. Let's go ahead and figure out what this c value is. So this is going to be our second step because remember we were told that f prime of 1 was 2. So we know our derivative value is 2 when our x value is 1. So we'd have 2 equals 1 half. I'm simply putting x equal to 1 plus 2 thirds plus c. If I solve that I get 5 sixths. So what we have so far is that f prime of x is equal to x squared over 2 plus 2 thirds x to the 3 halves plus 5 sixths. But now we have to do the anti-derivative again so we can arrive back at our original function. So our third step is that we will arrive at our original function by doing the anti-derivative of what I had for my first derivative. That first derivative was 1 half x squared plus 2 thirds x to the 3 halves plus 5 sixths. Remember we add 1 to our exponents. The 1 half stays as our coefficient adding 1 to the exponent and then dividing by it we get x cubed over 3 plus now our 2 thirds stays. When we add 1 to our exponent here in the middle we get 5 halves but then we have to multiply by the reciprocal 2 fifths plus last term anti-derivative is just 5 sixths x. And now we have a new c to find. Let me simplify this a little bit. Going back to the other initial condition we were given we were told that f of 1 was equal to 1 so we can use that fact to now find this particular c value. So the function value is 1 when x equals 1 so in place of these x's I'm going to substitute 1 so I have 1 sixth plus 4 fifteenths plus 5 sixths plus c. So you get a c value of negative 4 fifteenths. You typically would see these as fractions as opposed to decimals. That means therefore our final answer for our function is that our function is 1 sixths x to the third plus 4 fifteenths x to the fifth half plus 5 sixths x minus 4 fifteenths. Now one side note, if by chance initial conditions had not been given and you were simply asked to solve this differential equation one thing you would have to watch as you walked through was how to denote your constants because you cannot have two c's. This first c came from doing the antiderivative of the second derivative to arrive at the first derivative. If by chance these initial conditions were not given you would be working through another antiderivative using this as c instead of having a number there. So when you went to find the antiderivative of c over here on this next step what you would have it would have been right in here instead of 5 sixths x you would have had c times x right there because you didn't know yet c was 5 sixths because you were not given initial conditions. If that was the case you can't you see again you'd have to pick a different letter pick anything I usually pick d. A lot of textbooks you'll see these denoted as c1 and c2 which is fine as well but that's just one thing to look for if by chance you were not given initial conditions and yet had to solve that differential equation you just have to be careful how you denote your different constants.