MATH 308 Lecture 36

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Lecture Notes

Section 7.9

Solving nonhomogeneous systems:

• Undetermined coefficients
• Variation of parameters
• Laplace transforms

$X(t)=X_p(t)+X_h(t)$

where $X_p(t)$ is a particular solution and $X_h(t)$ is the general solution to the homogeneous form problem.

Exercise 1a: Undetermined Coefficients

$X^{\prime }(t)=\left[\begin{array}{cc}1& 1\\ 4& -2\end{array}\right]X+\left[\begin{array}{c}{\mathrm{e}}^{-2t}\\ -2{\mathrm{e}}^t\end{array}\right]$

Solution to homogeneous system is $X_h(t)=c_1{\mathrm{e}}^{-3t}\left[\begin{array}{c}1\\ -4\end{array}\right]+c_2{\mathrm{e}}^{2t}\left[\begin{array}{c}1\\ 1\end{array}\right]$

Guess for particular solution:

$\begin{array}{rl}{X}_p(t)& ={\mathrm{e}}^{-2t}\overrightarrow{v}+{\mathrm{e}}^t\overrightarrow{u}\\ & =\left[\begin{array}{c}{v}_1{\mathrm{e}}^{-2t}+u_1{\mathrm{e}}^t\\ v_2{\mathrm{e}}^{-2t}+u_2{\mathrm{e}}^t\end{array}\right]\end{array}$

Solve for derivative:

$\begin{array}{rl}{X_p}^{\prime }(t)& =-2{\mathrm{e}}^{-2t}\overrightarrow{v}+{\mathrm{e}}^t\overrightarrow{u}\\ A{X}_p+G(t)& ={\mathrm{e}}^{-2t}\left[\begin{array}{cc}1& 1\\ 4& -2\end{array}\right]\overrightarrow{v}+{\mathrm{e}}^t\left[\begin{array}{cc}1& 1\\ 4& -2\end{array}\right]\overrightarrow{u}+{\mathrm{e}}^t\left[\begin{array}{c}0\\ -2\end{array}\right]+{\mathrm{e}}^{-2t}\left[\begin{array}{c}1\\ 0\end{array}\right]\\ -2{\mathrm{e}}^{-2t}\overrightarrow{v}+{\mathrm{e}}^t\overrightarrow{u}& ={\mathrm{e}}^{-2t}\left[\begin{array}{c}{v}_1+v_2\\ 4v_1-2v_2\end{array}\right]+{\mathrm{e}}^t\left[\begin{array}{c}{u}_1+u_2\\ 4u_1-2u_2\end{array}\right]+{\mathrm{e}}^t\left[\begin{array}{c}0\\ -2\end{array}\right]+{\mathrm{e}}^{-2t}\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \end{array}$

Solve system term-wise:

Coefficients of ${\mathrm{e}}^{-2t}$:

$\{\begin{array}{l}{v}_1+v_2+1=-2v_1\\ 4v_1-2v_2=-2v_2\end{array}$

From this we can determine that $\overrightarrow{v}=⟨0,-1⟩$

Coefficients of ${\mathrm{e}}^t$:

$\{\begin{array}{l}{u}_1+u_2=u_1\\ 4u_1-2u_2-2=u_2\end{array}$

From this we can determine that $\overrightarrow{u}=⟨\frac{1}{2},0⟩$

Exercise 1b: Variation of Parameters

$X^{\prime }(t)=\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right]X+\left[\begin{array}{c}{\mathrm{e}}^t\\ t\end{array}\right]$

Solution to homogeneous system is $X_h(t)=c_1{\mathrm{e}}^t\left[\begin{array}{c}1\\ 1\end{array}\right]+c_2{\mathrm{e}}^{-t}\left[\begin{array}{c}1\\ 3\end{array}\right]$

Guess for particular solution:

$\begin{array}{rl}{X}_p(t)& =t{\mathrm{e}}^t\overrightarrow{u}+t\overrightarrow{v}+\overrightarrow{w}+{\mathrm{e}}^t\overrightarrow{x}\\ & =\left[\begin{array}{c}{u}_{1}t{\mathrm{e}}^t+t{v}_1+w_1+x_1{\mathrm{e}}^t\\ u_{2}t{\mathrm{e}}^{-t}+t{v}_2+w_2+x_2{\mathrm{e}}^t\end{array}\right]\end{array}$

Find derivative and set equal to substituted RHS

$\begin{array}{rl}{X_p}^{\prime }(t)& =({\mathrm{e}}^t+t{\mathrm{e}}^t)\overrightarrow{u}+\overrightarrow{v}\\ A{X}_p(t)+G(t)& =t{\mathrm{e}}^t\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right]\overrightarrow{u}+t\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right]\overrightarrow{v}+\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right]\overrightarrow{w}+{\mathrm{e}}^t\left[\begin{array}{c}1\\ 0\end{array}\right]+t\left[\begin{array}{c}0\\ 1\end{array}\right]\\ {\mathrm{e}}^t\overrightarrow{u}+\overrightarrow{v}& ={\mathrm{e}}^t\left[\begin{array}{c}2u_1-u_2\\ 3u_1-2u_2\end{array}\right]+t\left[\begin{array}{c}2v_1-v_2\\ 3v_1-2v_2\end{array}\right]+\left[\begin{array}{c}2w_1-w_2\\ 3w_1-2w_2\end{array}\right]+{\mathrm{e}}^t\left[\begin{array}{c}1\\ 0\end{array}\right]+t\left[\begin{array}{c}0\\ 1\end{array}\right]\\ & ⋮\\ X_p(t)& =t{\mathrm{e}}^t\left[\begin{array}{c}\frac{3}{2}\\ \frac{3}{2}\end{array}\right]+t\left[\begin{array}{c}1\\ 2\end{array}\right]+\left[\begin{array}{c}0\\ -1\end{array}\right]+{\mathrm{e}}^t\left[\begin{array}{c}\frac{1}{2}\\ 0\end{array}\right]\end{array}$

Therefore, the general solution is the sum $X(t)=X_p(t)+X_h(t)$:

$X(t)=t{\mathrm{e}}^t\left[\begin{array}{c}\frac{3}{2}\\ \frac{3}{2}\end{array}\right]+t\left[\begin{array}{c}1\\ 2\end{array}\right]+\left[\begin{array}{c}0\\ -1\end{array}\right]+{\mathrm{e}}^t\left[\begin{array}{c}\frac{1}{2}\\ 0\end{array}\right]+\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right]X+\left[\begin{array}{c}{\mathrm{e}}^t\\ t\end{array}\right]$