 Do you remember the differential equation for the oscillating mass? This is a second-order differential equation with constant coefficients. The perturbation function is zero. That means we only have to find out the homogeneous solution. And we do that with the exponential ansatz we just learned. Let's take the fast way and directly write down the characteristic equation. We expect a quadratic equation because we have a second-order differential equation. The second derivative is preceded by the coefficient one, so we just write lambda squared. Then the coefficient in front of the first derivative. Since the first derivative is missing in the differential equation, the lambda term is absent as well. Next up is the coefficient d over m that is in front of the searched function. This coefficient stands alone in the characteristic equation. Altogether the characteristic equation reads lambda squared plus d over m is equal to zero. For this equation we don't even need the quadratic formula. We get the solution directly if we first bring d over m to the other side and then take the square root. Consider that the inverse of squaring gives two solutions, a positive and a negative square root. Also we have an interesting case here when the square root of a negative number is taken. Square root of a negative number is not a real number, but an imaginary number. Do you remember what that means? We expect that the system must oscillate. Even if you don't know imaginary or complex numbers yet, you can split the term inside the square root into a product of minus one and d over m. According to the square root laws, you can split this product into two square roots. Square root of minus one is defined as the imaginary unit, a number which we denote by the letter i. That's all you need to know about imaginary numbers. Let us denote square root of d over m shortly as omega. If we insert the lambda values we have just found into the exponential ansatz, we get the general solution for the considered differential equation. This solution seems very abstract, but I will show you that this solution corresponds to an oscillation. Before that, let us determine the unknown constants c1 and c2 with initial conditions for our problem. For example, we could have observed that at the time t equals zero, the displacement of the spring was one. The spring was displaced to the maximum, so the initial condition is y of zero is equal to one. Insert this condition into the solution to determine c1. e to the power of zero is one, rearrange for c1 and you get c1 is equal to one minus c2. Next step is to determine the unknown constant c2 using the second initial condition. For this we use the fact that at t equals zero, the velocity of the mass was zero. In physics you learn that velocity corresponds to the first time derivative of the displacement. So our second initial condition is given by y' of zero is equal to zero. So let's differentiate our general solution with respect to time t. The factor in front of t becomes a factor in front of the exponential function. And then we insert the second initial condition into the derivative. The exponential functions become one and the factor i omega cancels out. Rearranging for c2 we find out that c2 is equal to c1. So we know that c2 must be equal to c1. Nice. Let's replace c2 with c1 to determine concrete value for the constants. The equation results in c1 is equal to one half. Since c1 and c2 are equal, c2 must be also equal to one half. Insert one half into our general solution. Now let's find out what this abstract solution has to do with oscillations. For this we get our friend Euler to help us, who tells us his famous Euler formula. e to the power of i omega t is equal to cosine of omega t plus i times sine of omega t. This relation tells us how the complex exponential function is related to cosine and sine functions. So the first complex exponential in our solution becomes cosine and sine with positive omega t and the second complex exponential becomes cosine and sine with negative omega t. We can omit the minus sign in the argument of the cosine function because cosine is symmetric. That means it has the same value for arguments x and minus x. The sine function on the other hand is anti-symmetric. Therefore we cannot omit the minus sign in the argument but rather pull it out in front of the sine function. Very nice. The complex sine function drops out and the cosine can be summed up. And this is our final solution. As you can see, the displacement y changes periodically with time. The mass attached to the spring oscillates back and forth and the oscillation is described by the cosine function.