# Undetermined coefficients ```{topic} Undetermined coefficients If a nonhomogeneous linear second-order ODE has constant coefficients \begin{equation} y'' + a y' + b y = r(x) \end{equation} we can "guess" a solution as follows: | $r$ | $y_{\rm p}$ | |--------------------------------|---------------------------------------------------------| | $k e^{\gamma x}$ | $k e^{\gamma x}$ | | $k x^n (n=0,1,\ldots)$ | $k_n x^n + k_{n-1} x^{n-1} + \cdots + k_1 x + k_0$ | |$k e^{\alpha x} \cos{\omega x}$ | $e^{\alpha x}(k_1 \cos{\omega x} + k_2 \sin{\omega x})$ | |$k e^{\alpha x} \sin{\omega x}$ | $e^{\alpha x}(k_1 \cos{\omega x} + k_2 \sin{\omega x})$ | - If *r* has multiple terms, add guesses for each together. - If the guess for $y_{\rm p}$ is a homogeneous solution $y_{\rm h}$, multiply by *x* unless this solution is a repeated root - then multiply by $x^2$! ``` For example, to solve: \begin{equation} y'' + y = 0.001 x^2, \quad y(0) = 0, \quad y'(0) = 1.5 \end{equation} Use the following steps: 1. Find $y_{\rm h}$: \begin{align} y_{\rm h}'' + y_{\rm h} &= 0 \\ \lambda^2 + 1 &= 0 \end{align} so $\lambda = \pm i$ and \begin{equation} y_{\rm h} = c_1 \cos x +c_2 \sin x \end{equation} 2. Guess $y_{\rm p}$ and solve for undetermined coefficients: \begin{align} y_{p} &= k_2 x^2 + k_1 x + k_0 \\ y_{\rm p}' &= 2k_2 x + k_1 \\ y_{\rm p}'' &= 2k_2 \end{align} So \begin{equation} y_{\rm p}'' + y_{p} = 2 k_2 + k_2 x^2 + k_1 x + k_0 = 0.001x^2 \end{equation} Equating coefficients of like powers of *x* gives the system \begin{align} k_2 &= 0.001 \\ k_1 &= 0 \\ 2k_2 + k_0 &= 0 \\ \end{align} which gives $k_0 = -0.002$. 3. Combine solutions and solve IVP. The general solution and its first derivative are: \begin{align} y &= y_{\rm h} + y_{\rm p} = c_1 \cos x + c_2 \sin x + 0.001x^2 - 0.002 \\ y' &= -c_1 \sin x + c_2 \cos x + 0.002x \end{align} Substituting in the first boundary condition: \begin{align} y(0) &= c_1 - 0.002 = 0\\ y'(0) &= c_2 = 1.5 \end{align} Hence, $c_1 = 0.002$, $c_2 = 1.5$, and \begin{equation} y = 0.002 \cos x + 1.5 \sin x + 0.001x^2 - 0.002 \end{equation} --- As a more complex example, let's find the form of the particular solution for: \begin{equation} y'' + 3y' + 2.25y = -10e^-1.5x + \cos x \end{equation} 1. Find $y_{\rm h}$: \begin{align} y_{\rm h}''+3y_{\rm h}'+2.25y_{\rm h} &= 0 \\ \lambda^2 +3\lambda+2.25 = 0 \\ (\lambda+1.5)^2 &= 0 \\ \end{align} so $\lambda = -1.5$. This is a real repeated root, so \begin{equation} y_{\rm h} = (c_1 + c_2 x)e^{-1.5x} \end{equation} 2. Guess $y_{\rm p}$: We should add guesses for both terms in *r*. Our first term is an exponential, but it needs to be multiplied by $x^2$ because it is the same as the homogeneous solution, which is a real repeated root. Our second term is a sum of cosine and sine. \begin{align} y_{\rm p} = k_1 x^2 e^{-1.5 x} + (k_2 \cos x +k_3 \sin x) \end{align} ## Skill builder problems 1. Solve: \begin{equation} y'' + 3y' + 2y = 30e^{2x}, \quad y(0) = 1, \quad y'(0) = 0 \end{equation} ```{solution} 1. Find homogeneous solution $y_{\rm h}$: \begin{align} y_{\rm h}'' + 3y_{\rm h}' + 2y_{\rm h} &= 0 \\ \lambda^{2} + 3\lambda + 2 &= 0 \\ (\lambda + 1)(\lambda + 2) &= 0 \end{align} so $\lambda_1 = -1$ and $\lambda_2 = -2$. Then, \begin{equation} y_{\rm h} = c_1 e^{x} + c_2 e^{-2x} \end{equation} 2. Find particular solution $y_{\rm p}$: \begin{align} y_{\rm p} &= ke^{2x} \\ y_{\rm p}' &= 2ke^{2x} \\ y_{\rm p}'' &= 4ke^{2x} \end{align} Combine and solve for *k*: \begin{align} 4ke^{2x} e^{2x} &+ 6ke^{2x} e^{2x} + 2ke^{2x} e^{2x} = 30e^2x e^{2x} \\ 12k &= 30 \end{align} so $k = 5/2$ and \begin{equation} y_{\rm p} = \frac{5}{2} e^{2x} \end{equation} 3. Combine and apply initial conditions: The general solution and its first derivative are \begin{align} y &= c_1 e^{-x} + c_2 e^-{2x} + \frac{5}{2} e^{2x} \\ y' &= -c_1 e^{-x} - 2c_2 e^{-2x} + 5e^{2x} \end{align} Plugging in initial conditions \begin{align} y(0) &= c_1 + c_2 + \frac{5}{2} = 1 \\ y'(0) &= -c_1 - 2c_2 + 5 = 0 \end{align} This is a system of linear equations: \begin{align} c_1 + c_2 &= -\frac{3}{2} \\ c_1 + 2c_2 &= 5 \\ \end{align} that can be solved to give $c_1 = -8$ and $c_2 = 13/2$. The final solution is: \begin{equation} y = -8e^{-x} + \frac{13}{2} e^{-2x} = \frac{5}{2} e^{2x} \end{equation} ``` 2. Solve: \begin{equation} y'' + 4y = 16 \cos 2x , \quad y(0), \quad y'(0) = 0 \end{equation} ```{solution} 1. Find homogeneous solution $y_{\rm h}$: \begin{align} y_{\rm h}'' + 4y_{\rm h} &= 0 \\ \lambda^{2} + 4 &= 0 \\ \end{align} so $\lambda_{1,2} = \pm 2i$ and \begin{equation} y_{\rm h} = c_1 \cos 2x + c_2 \sin 2x \end{equation} 2. Find the particular solution $y_{\rm p}$: \begin{align} y_{\rm p} &= x(k_1 \cos 2x + k_2 \sin 2x) \\ y_{\rm p}' &= x(-2k_1 \sin 2x + 2k_2 \cos 2x) + (k_1 \cos 2x + k_2 \sin 2x) \\ y_{\rm p}'' &= x(-4k_1 \cos 2x - 4k_2 \sin 2x) + 2(-2k_1 \sin 2x + 2k_2 \cos 2x) \end{align} Combine: \begin{equation} y_{\rm p}'' + 4y_{\rm p} = 2(-2k_1 \sin 2x + 2k_2 \cos 2x) = 16\cos 2x \end{equation} Comparing terms gives $k_1 = 0$ and $k_2 = 4$. 3. Combine and apply initial conditions: The general solution and its first derivative are \begin{align} y &= c_1 \cos 2x + c_2 \sin 2x + 4x \sin 2x \\ y' &= -2c_1 \sin 2x + 2c_2 \cos 2x + 4x(2 \cos 2x) + 4 \sin 2x \end{align} Plugging in the initial conditions: \begin{align} y(0) &= c_1 = 0 \\ y'(0) &= 2c_2 = 0 \end{align} gives $c_1 = 0$ and $c_2 = 0$. The final solution is: \begin{equation} y = 4x \sin 2x \end{equation} ``` 3. Solve: \begin{equation} y'' - y' - 12y = 144x^3 + \frac{25}{2}, \quad y(0) = 5, \quad y'(0) = -1/2 \end{equation} ```{solution} 1. Find homogeneous solution $y_{\rm h}$: \begin{align} y_{\rm h}'' - y_{\rm h}' - 12y_{\rm h} &= 0 \lambda^2 - \lambda - 12 &= 0 (\lambda - 4)(\lambda + 3) &= 0 \end{align} so $\lambda_1 = 4$, $\lambda_2 = -3$, and \begin{equation} y_{\rm h} = c_1 e^{4x} + c_2 e^{-3x} \end{equation} 2. Find the particular solution $y_{\rm p}$: \begin{align} y_{\rm p} &= k_1 x^3 + k_2 x^2 + k_3 x + k_4 \\ y_{\rm p}' &= 3k_1x^2 + 2k_2x + k_3 \\ y_{\rm p}'' &= 6k_1x + 2k_2 \end{align} Substitute in ODE and combine: \begin{align} & y_{\rm p}'' - y_{\rm p}' - 12y_{\rm p} \\ &= (6k_1x + 2k_2) - (3k_1x^2 + 2k_2x + k_3) - 12(k_1x^3 + k_2x^2 + k_3x + k_4) \\ &= -12k_1x^3 + (-3k_1 - 12k_2)x^2 + (6k_1 - 2k_2 - 12k_3)x + (2k_2 - k_3 - 12k_4) \end{align} Comparing coefficients of powers of *x* gives: \begin{align} -12k_1 &= 144 \\ -3k_1 - 12k_2 &= 0 \\ 36 - 12k_2 &= 0 \\ 6k_1 - 2k_2 - 12k_3 &= 0 \\ 2k_2 - k_3 - 12k_4 &= \frac{25}{2} \end{align} Solving this system of linear equations gives $k_1 = -12$, $k_2 = 3$, $k_3 = -13/2$, and $k_4 = 0$, so: \begin{equation} y_{\rm p} = -12x_3 + 3x^2 - \frac{13}{2}x \end{equation} 3. Combine and apply boundary conditions: The general solution and its first derivative are \begin{align} y &= c_1 e^{4x} + c_2 e^{-3x} - 12x^3 + 3x^2 - \frac{13}{2}x \\ y' &= 4c_1 e^{4x} - 3c_2e^{-3X} - 36x^2 + 6x - \frac{13}{2} \end{align} Plugging in the initial conditions: \begin{align} y(0) &= c_1 + c_2 = 5 \\ y'(0) &= 4c_1 - 3c_2 - \frac{13}{2} = -\frac{1}{2} \end{align} Solve using matrices: \begin{align} \begin{bmatrix} 1 & 1 & 5 \\ 4 & -3 & 6 \end{bmatrix} \begin{matrix} \vphantom{R_1} \\ -4 R_1 \end{matrix} &\to \begin{bmatrix} 1 & 1 & 5 \\ 0 & -7 & 14 \end{bmatrix} \begin{matrix} \vphantom{R_1} \\ \div -7 \end{matrix} \\ &\to \begin{bmatrix} 1 & 1 & 5 \\ 0 & 1 & 2 \end{bmatrix} \begin{matrix} -R_2 \\ \vphantom{R_1} \end{matrix} \\ &\to \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & 2 \end{bmatrix} \end{align} so $c_1 = 3$ and $c_2 = 2$. The final solution is: \begin{equation} y = 3e^{4x} + 2e^{-3x} - 12x^3 + 3x^2 - \frac{13}{2}x \end{equation} ```