Solving Second-Order Linear Homogeneous Constant Coefficient Ordinary Differential Equations

If that title seems like a mouthful, it's because our studies so far have taught us how to solve just one very specific kind of ordinary differential equation (ODE). If you're unsure of how to determine if your equation fits the bill, check out my last post on classifying ODEs. To put it shortly, your equation should take the form ay'' + by' + cy = 0, where a, b, c are constants.

To start, find the characteristic polynomial. This is just aα² + bα + c = 0. Your professor may have used r instead of alpha. Now, solve for α using the quadratic formula, α = (-b ± √(b² - 4ac))/(2a). An astute reader might notice at this point that our problem branches off into three possible situations.

First possibility is that b² - 4ac > 0. If this occurs, the term under the radical is positive, and you receive 2 real solutions for α, we'll call them α₁ and α₂. Your solution to the ODE will then take the form y(x) = c₁e^α₁x + c₂e^α₂x.

The second situation is that b² - 4ac = 0. In this case, the ± term in your quadratic equation vanishes, and you therefore get just one root, which we'll call α. Your general solution will now take the form y(x) = c₁e^αx + c₂xe^αx.

Finally, you might run into the situation that b² - 4ac < 0, which means that the equation has imaginary roots equal to -b/(2a) ± i√(4ac - b²)/(2a). An imaginary number has two distinct parts, the real part and the imaginary part, written in the form β+iγ (i is √-1). In this case,  β = -b/(2a), and γ = √(4ac - b²)/(2a). And the general solution takes the form y(x) = c₁e^βxcos(γx)+ c₂e^βxsin(γx).

c₁ and c₂ represent any constant, and these cannot be found specifically without being given initial conditions - that is, a value for y(0) and y'(0). If you are given initial conditions, you must take some additional steps to determine c₁ and c₂. Plug in 0 for x and set the resulting expression equal to y(0). This will give you one equation involving c₁ and c₂. Then differentiate y(x), plug in 0 for x, and set the resulting expression equal to y'(0). You will then have a system of two equations involving c₁ and c₂. Use your choice of methods (substitution or elimination) to find c₁ and c₂.

These are just the answers! Your professor may want you to show many more steps than what I have shown here. And as always, if you have any questions, come to MASH. And if there's anything specific that you'd like to see on the blog, leave a comment. You don't need to have a blogspot account to comment.

Happy calculating,
Rachel