# Component-wise Undetermined Coefficients

Every resource I have seen on undetermined coefficients for inhomogeneous systems uses only examples in which the components of the forcing term have a common replication under differentiation. Thus the solution techniques suggest guesses of forms such as $$\vec{x_p} = \vec{a}e^{2t}$$ or $$\vec{x_p} = \vec{a}t^2 + \vec{b}t + \vec{c}$$, where each component of one of the undetermined vectors plays the same role (i.e., $$b_1$$ and $$b_2$$ are the coefficients of the $$t$$ terms).

Can undetermined coefficients also be used to solve systems with forcing terms in which each component has a different replication under differentiation, by making the usual guesses component-wise? For example, if the forcing term is $$\begin{bmatrix} .5t \\ \sin(3t) \end{bmatrix}$$, the guess would be $$\vec{x_p} = \begin{bmatrix} at + b \\ c\cos(3t) + d\sin(3t) \end{bmatrix}$$. For each component, the guess is made which would be made if that component were the forcing term in a second-or-higher-order single ODE. Is this a legitimate way to generalize undetermined coefficients for systems, or does the technique only work if all components share a common replication under differentiation?

• @Moo So does what I suggested work as a shortcut, or do you have to solve the two systems separately? Feb 24 '20 at 5:09

## 1 Answer

The first step is to come as close as possible to decoupling the variables. For example, consider $$\begin{bmatrix} \dot{x_1}\\ \dot{x_2}\end{bmatrix}=\begin{bmatrix}5&4\\-2&1 \end{bmatrix} \begin{bmatrix}x_1\\x_2 \end{bmatrix} +\begin{bmatrix} f(t)\\g(t)\end{bmatrix}$$, The matrix $$\begin{bmatrix} 5&4\\-2&1\end{bmatrix}$$ has eigenvalues $$3\pm 2i$$. There exists a real non-singular matrix $$P$$ such that $$P^{-1}\begin{bmatrix} 5&4\\-2&1\end{bmatrix}P=\begin{bmatrix}3&-2\\2&3 \end{bmatrix}.$$ Make the change of variable $$\begin{bmatrix} x_1\\x_2\end{bmatrix} =P\begin{bmatrix} u_1\\u_2\end{bmatrix}$$ Then $$\begin{bmatrix}\dot{u_1}\\ \dot{u_2}\end{bmatrix}=\begin{bmatrix}3&-2\\2&3\end{bmatrix} \begin{bmatrix}u_1\\u_2\end{bmatrix} +P^{-1}\begin{bmatrix}f(t)\\g(t)\end{bmatrix}$$.Let us suppose, for the sake of illustration, that $$P^{-1}\begin{bmatrix}f(t)\\g(t)\end{bmatrix}=\begin{bmatrix}2t+1\\5\sin(2t)\end{bmatrix}$$ which gives us $$\begin{bmatrix}\dot{u_1}\\ \dot{u_2}\end{bmatrix}=\begin{bmatrix}3&-2\\2&3\end{bmatrix} \begin{bmatrix}u_1\\u_2\end{bmatrix} +\begin{bmatrix}2t+1\\5\sin(2t)\end{bmatrix}$$. Our second step is to use the method of undetermined coefficients to find particular solutions $$u_{1p}$$ and $$u_{2p}$$ of these equations. Your question is what expressions to use. Both for $$u_1$$ and $$u_2$$ we take any expression that appears in either one of the equations. So we set $$u_{1p}=at+b+r\cos(2t)+s\sin(2t),$$ $$u_{2p}=a’t+b’+r’\cos(2t)+s’\sin(2t).$$. Substitute these expressions for $$u_{1p}$$ and $$u_{2p}$$ into $$\dot{u_1}-(3u_1-2u_2)$$and $$\dot{u_2}-(2u_1+3u_2)$$ respectively and obtain expressions $$(…)t+(…)+(…)\cos(2t)+(…)\sin(2t)$$. Set each (…) equal to 0, giving 8 linear equations in 8 unknowns. Solve these. The complementary solutions, i.e. the solutions of the homogeneous equations for $$u_1$$ and $$u_2$$ are $$u_{1c}=Ce^{3t}\cos(2t)+De^{3t}\sin(2t), u_{2c}=-De^{3t}\cos(2t)+Ce^{3t}\sin(2t)$$ The general solution for $$u_1$$ and $$u_2$$ is $$u_1=u_{1c}+u_{1p}, u_2=u_{2c}+u_{2p}$$. From initial conditions on $$x_1$$ and $$x_2$$ , we multiply by $$P^{-1}$$ to obtain initial conditions on $$u_1$$ and $$u_2$$, which we use to find $$C$$ and $$D$$. Having completely found $$u_1$$ and $$u_2$$,we multiply by $$P$$ to find $$x_1$$ and $$x_2$$.

• What is the point of partially decoupling the ODE? I thought decoupling was only used because systems with diagonal matrices are trivial. Could this step be omitted? Feb 27 '20 at 19:42
• The point of partial decoupling, which is the best we can do if we wish to stay inside the real numbers, is to obtain a block-diaonal form in which the blocks have a very special form, i.e. a rotation matrix, which allows us to write the solution of the corresponding homogeneous equation, the so-called "complementary" solution, in a particularly nice form. Feb 27 '20 at 20:48