As the second topic in chapter one, I will consider the so-called initial value problem. Quite often, interesting mathematical modeling leads to not only a differential equation but also some side conditions on the unknown function or its derivatives. We now introduce one such problem, okay? For example, the problem goes like this one, okay? Find the solution to an nth order ordinary differential equation, right? This is equation four, okay? Satisfying initial conditions, say, n initial conditions, given by y of x naught = y naught y prime of x naught 0 = y sub 1, n- 1 derivative of y of x naught = y sub n- 1, okay? How many initial conditions do we have here? One, two, dot dot dot and n, right? The number of initial conditions is n, okay? And in the order of differential equation is also n, they coincide, okay? So we are looking for a solution of this differential equation, satisfying additional n initial conditions given by this on some interval I containing the so called initial point x naught, where the x naught, y naught, y1, and y sub n- 1 are given constants, okay? We call such a problem as an initial value problem or sometimes people call it as a Cauchy problem, okay? Also call it as Cauchy problem, okay? Let's look at a couple simple examples of initial value problems. As a first example, Let's show that y = 1 over x squared + c is a general solution of the first order differential equation, which is nonlinear, y prime + 2x y squared = 0, okay? Then find the solution of the initial value problem for this differential equation, satisfying the initial condition y(0) = one-half, okay? So here you have y = 1 over x squared + c, is the same as x squared + c to the -1, right? So its derivative is, y prime = negative x squared + c to the negative 2 times, derivative of inside that is equal to 2x, right? On the other end, what is 2xy squared? This is = 2x times, square of this will be, x squared + c to the -2, all right? Add these two then, left-hand side is exactly y prime + 2xy squared right? Over to right hand side, right hand side is -2x times x squared + c to the -2. Here, you have 2x times x squared + c to the -2, so that canceled, makes 0, right? So this is really a solution to given differential equation and since the differential equation is of order 1. And here you have 1 arbitrary constant, 1 free parameter, so this is a general solution of that differential equation, okay? Secondly, we are required to find the solution of that initial value problem, satisfying the differential equation, and the initial condition y(0) = one-half right? So starting from this one and try to find the constant c so that this extra initial condition is satisfied, okay? So how much is one-half must be y(0), from this one, when x = 0, this is = c to the-1, right? What does this then mean, that means c = 2, right? So therefore your solution, our solution we are looking for is x squared + 2 to the -1, okay? That's a solution to this initial value problem, okay? Here is one another question, does this differential equation has a singular solution? Or let me ask in this way, does this differential equation have a trivial solution? What is a trivial solution? Function which is identically 0, if it satisfies a differential equation, then we call it a trivial solution, okay? y = to identically 0, satisfied this differential equation trivial, okay? So that the given differential equation has a trivial solution, right? Can you obtain the trivial solution by specifying the values of c in this expression? This is simply impossible, right? y = 1 over x squared + c will never be identically = 0 for no choice of c, right? So that the trivial solution is also singular solution of a given differential equation, right? As a second example, Show that y = sine x- cosine of x is a solution of the initial value problem, y double prime + y = 0. Satisfy y(0) = -1 and y prime of 0 = 1, okay? Note that the differential equation is a linear second order, and we have two initial conditions, right? So now, we have y = sine of x- cosine x, y prime = cosine of x + sine of x, y double prime = -sine of x + cosine of x, right? From this make y double prime + y, add this first and third equation. This sine of x, and -sine x, cancels out, right? This -cosine x, this positive cosine x, canceled out, right? So really, y double prime + y = 0 when y = sine x- cosine x, right? That means this is a solution to the given differential equation, right? Is it a solution to the initial value problem, how much is y(0)? When x = 0, sine 0 = 0, cosine 0 = 1, so y of 0 will be -1. How much is y-prime of 0, from this second equation, when x = 0, cosine 0 is 1, sine 0 is 0. So y prime of 0, that is positive 1, is the second initial condition. So that the given function, sine x- cosine x, satisfies both differential equation and the initial conditions, we confirmed it.