is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Nonhomogeneous Equations with Constant Coefficients

This is very similar to the last post. 

MathJax TeX Test Page Let's say you have $Ay'' + By' + Cy = f(x)$, you have to do two things. First, you find solution to the associated homogeneous differential equation. Why? Because let's say w is a particular solution to the nonhomogeneous differential equation, and y is the set of solutions to the homogeneous dif eq. The set of all solutions to the nonhomogeneous differential equation is y + w. That's because $L(y + w) = L(y) + L(w) = 0 + f(x) = f(x)$.

Second, you find the particular solution. I will demonstrate how to do this with an example.


MathJax TeX Test Page $$y'' + 4y' + 4y = 3x + 2$$ First, we solve the complementary y, or the homogeneous equation. $$y_c = c_1e^{-2x} + c_2xe^{-2x}$$ Now, we look at all of the terms in f(x) and all of its derivatives, we get that the particular equation equals $y_p = Ax + B$. We plug that in and we get $$4A + 4Ax + 4B = 3x + 2 \to 4Ax = 3x \to A = \frac{3}{4}$$ $$4A + 4B = 2 \to A + B = \frac{1}{2} \to B = -\frac{1}{4}$$ Our final equation is $y = y_c + y_p$ $$y = c_1e^{-2x} + c_2xe^{-2x} + \frac{3}{4}x - \frac{1}{4}$$

Example- Repeat Roots

MathJax TeX Test Page $$y''' - y'' - 12y' = x - 2xe^{-3x}$$ First, we find the homogeneous solution. $$r^3 - r^2 - 12r = 0 \to r(r + 3)(r-4) = 0$$ $$y_c = c_1 + c_2e^{-3x} + c_3e^{4x}$$ Ignore $y_c$ for now, $y_p$ would be $Ax + B + Ce^{-3x} + Dxe^{-3x}$. However, let's factor in the repeated roots. Because there's already a constant coefficient, we multiply that term by x. However, there's already an x term! So we multiply that by x as well. Same thing with $e^{-3x}$. That's already in the $y_c$ solution. So, we multiply the $e^{-3x}$ terms by x, but that conflicts with $xe^{-3x}$, so we multiply that by x as well. Our new equation for $y_p$ is $$Ax^2 + Bx + Cxe^{4x} + Dx^2e^{-3x}$$

That's pretty much it. It's very similar to the homogeneous equation, there's a just an additive factor that pushes all of the solutions up from 0 to f(x).

David Witten

Euler-Cauchy Equations

Homogeneous Equations with Constant Coefficients