 In general, we can write a linear differential equation in the form Ly equals h of x, where l is a linear differential operator. If h of x equals 0, we can solve it by finding the roots of the characteristic polynomial by our theorem, if l is our linear differential operator and r is a root, then ce to power rx is a solution. We won't worry about how to find the roots, assuming we can find them algebraically, numerically, or through wild guesses and wishful thinking, but there are some other problems that arise when trying to use this approach to solve differential equations. So one of the problems originates with differential equations like y double prime plus y equals 0. In operator form, this is d2y plus y equals 0, and so we have characteristic polynomial d squared plus 1. The roots of this polynomial are complex numbers plus or minus i, and so the solutions are c1 e to the power ix and c2 e to the power minus ix. The what? What does it mean when we raise e to a complex number? To answer the question of what it means to raise e to a complex number, we'll have to introduce some new ideas. In 1722, the English mathematician Abraham de Moivre, wait a minute, de Moivre, is that really English? And well, yes, oh sure, de Moivre was born in France, but Christian fanatics forced de Moivre and thousands of others to flee across the Channel to Great Britain, where their influx added a new word to the English language, refugee. And so Abraham de Moivre should be considered an English mathematician. So this English mathematician Abraham de Moivre described the following relationship. For a and b real numbers, the expression e to the power a plus b i can be rewritten as e to the power a times cosine b plus i sine of b. This is useful enough that this is one of the few formulas that it's worth actually committing to memory. And so now our two apparently complex solutions, c one e to the power i x, well that's really c one cosine x plus i c one sine of x. Similarly, c two e to the power minus i x, well that'll be c two cosine minus x plus i c two sine minus x. But remember, cosine of minus x is just cosine of x, and sine of minus x is minus sine of x. And so this becomes, and so our second solution, c two e to minus i x is going to be c two cosine x minus i c two sine x. So when we add our two solutions, we can express that sum as c one plus c two cosine x plus i c one minus c two sine of x. Since we can absorb our constants, we can write this as follows. This c one plus c two, well that's just a constant, so we'll call it oh, c one, and i times c one minus c two, well that's really just another constant, and we'll call that c two. And so this is the general solution to the differential equation y double prime plus y equals zero. I'd check this if I were you. In general, this leads to the following. Suppose r equals a plus or minus b i are complex roots of the characteristic polynomial. Then e to power ax c one cosine b x plus c two sine b x are going to be solutions to the corresponding differential equation. So for example, let's consider the fourth order linear differential equation. c four y minus y is equal to zero. Rewriting this in operator notation. And so our characteristic polynomial is d to the fourth minus one. We can find the roots of the characteristic polynomial by setting it equal to zero and solving. And again, this is the one case out of billions where we could actually solve this problem algebraically. If we do that, we get four solutions plus or minus one plus or minus i. And so once we have the roots of our characteristic polynomial, we can translate these into components of our general solution, the roots plus or minus one because they're real give us the solutions c one e to power x and c two e to power minus x. Meanwhile, the two complex roots plus or minus i, those will give us trigonometric solutions. So remember that any pure imaginary number b i can be written as zero plus b i. So our solutions will look like e to power zero x times c three cosine x plus c four sine x. And while this is a perfectly good algebraic solution, you lose street credit among mathematicians if you don't remember that e to power zero x is one and since it'll be a factor of one, we don't actually need to write it down as part of our product. And so our general solution will be c one e to the x plus c two e to the minus x plus c three cosine x plus c four sine x. Or we might consider another example, rewriting our differential equation in operator form. In a kind and gentle universe, every quadratic equation would be factorable. Or if I were a kind and gentle math teacher, I would never give you any problems that required you to do anything difficult. You know where this is going, don't you? So remember, most quadratic equations can't be factored, and so most of them will have to be solved using the quadratic formula. So our characteristic equation is d squared plus d plus five equal to zero with solutions. Hint, this is not factorable, we'll have to use the quadratic formula. And they turn out to be complex roots. So the real part of the root will become an exponential e to the power minus one half x. The complex component without the I is going to become the argument of our cosine and sine functions. And our constants of anti-differentiation are going to show up as coefficients of cosine and sine, and so our general solution will look like this.