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

Homogeneous Equations with Constant Coefficients

MathJax TeX Test Page Let's say you have a homogeneous (equal to 0) differential equation $$Ay'' + By' + Cy = 0$$ Notice that because the only coefficients are constants, then y, y', and y'' all must be the same general form. What do we know that gives pretty much the same thing back when we differentiate it? $e^x$. We don't want to guess e^x, because $e^{2x}$could also work, so we guess $e^{rx}$.


MathJax TeX Test Page $$y'' + 3y' - 10y = 0$$ Guess $e^{rx}$ $$r^2(e^{rx}) + 3r(e^{rx}) - 10(e^{rx}) = (r^2 + 3r - 10)(e^{rx}) = 0$$ Because $e^x$ is never 0, we can divide through by that to get r = -2, 5. Therefore, our solutions are $e^{-2x}, e^{5x}$.

Here is an important thing to consider. If $L(y) = Ay'' + By' + Cy$, that means it's a linear function, because nothing's squared. Because it's linear, we can say $L(y_1 + y_2) = L(y_1) + L(y_2)$. Also $L(cy_1) = cL(y_1)$. By this same principle if $L(y_1) = 0$ and $L(y_2) = 0$, then $L(c_1y_1 + c_2y_2) = 0$. Therefore, our final solution is $$\boxed{c_1e^{-2x} + c_2e^{5x}}$$

Example 2- Imaginary

MathJax TeX Test Page Let's say you have this: $$y'' + y = 0$$ You turn this into: $$(r^2 + 1)e^{rx} = 0$$ $$r = \pm 1 \to e^{ix}, e^{-ix}$$ $$e^{ix} = \cos(x) + i\sin(x)$$ $$e^{-ix} = \cos(x) - \sin(x)$$ $$c_1(\cos(x) + i\sin(x)) + c_2(\cos(x) - i\sin(x))$$ $$(c_1 + c_2)\cos(x) + (c_1 - c_2)i\sin(x)$$ Side note: if $a + bi = 0$, then $a,b = 0$, because 1 and $i$ are linearly independent. This is important because if we calculate $L(y) = 0$, and $L(y_1) + L(iy_2) = 0$. We can remove the $i$, getting $L(y_1) + iL(y_2) = 0$. Because they are linearly independent, both $L(y_1)$ and $L(y_2)$ equal 0. Our solutions are now the same, except we can remove the $i$. So, the solutions are $$d_1\cos(x) + d_2\sin(x)$$
MathJax TeX Test Page Let's look at this: $$y'' + 4y' + 4y = 0$$ Then, $(r+2)^2$. We know we need two roots, but we only have one r. What do we do? We have $e^{-2x}$ twice. For the sake of simplicity, I have omitted the proof, but all we have to do is add x to one solution, so we get. For a thrice repeated root we add x and $x^2$. $$\boxed{c_1e^{-2x} + c_2xe^{-2x}}$$
David Witten

Nonhomogeneous Equations with Constant Coefficients

Picard Iterations