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

$X'(t)={\begin{bmatrix}1&1\\4&-2\end{bmatrix}}\,X+{\begin{bmatrix}\mathrm {e} ^{-2t}\\-2\mathrm {e} ^{t}\end{bmatrix}}$

Solution to homogeneous system is $X_{h}(t)=c_{1}\,\mathrm {e} ^{-3t}\,{\begin{bmatrix}1\\-4\end{bmatrix}}+c_{2}\,\mathrm {e} ^{2t}\,{\begin{bmatrix}1\\1\end{bmatrix}}$

Guess for particular solution:

${\begin{aligned}X_{p}(t)&=\mathrm {e} ^{-2t}\,{\vec {v}}+\mathrm {e} ^{t}{\vec {u}}\\&={\begin{bmatrix}v_{1}\,\mathrm {e} ^{-2t}+u_{1}\mathrm {e} ^{t}\\v_{2}\mathrm {e} ^{-2t}+u_{2}\mathrm {e} ^{t}\end{bmatrix}}\end{aligned}}$

Solve for derivative:

${\begin{aligned}X_{p}'(t)&=-2\mathrm {e} ^{-2t}\,{\vec {v}}+\mathrm {e} ^{t}\,{\vec {u}}\\A\,X_{p}+G(t)&=\mathrm {e} ^{-2t}\,{\begin{bmatrix}1&1\\4&-2\end{bmatrix}}\,{\vec {v}}+\mathrm {e} ^{t}\,{\begin{bmatrix}1&1\\4&-2\end{bmatrix}}\,{\vec {u}}+\mathrm {e} ^{t}\,{\begin{bmatrix}0\\-2\end{bmatrix}}+\mathrm {e} ^{-2t}\,{\begin{bmatrix}1\\0\end{bmatrix}}\\-2\mathrm {e} ^{-2t}\,{\vec {v}}+\mathrm {e} ^{t}\,{\vec {u}}&=\mathrm {e} ^{-2t}\,{\begin{bmatrix}v_{1}+v_{2}\\4v_{1}-2v_{2}\end{bmatrix}}+\mathrm {e} ^{t}\,{\begin{bmatrix}u_{1}+u_{2}\\4u_{1}-2u_{2}\end{bmatrix}}+\mathrm {e} ^{t}\,{\begin{bmatrix}0\\-2\end{bmatrix}}+\mathrm {e} ^{-2t}\,{\begin{bmatrix}1\\0\end{bmatrix}}\\\end{aligned}}$

Solve system term-wise:

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

${\begin{cases}v_{1}+v_{2}+1=-2v_{1}\\4v_{1}-2v_{2}=-2v_{2}\end{cases}}$

From this we can determine that ${\vec {v}}=\left\langle 0,-1\right\rangle$

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

${\begin{cases}u_{1}+u_{2}=u_{1}\\4u_{1}-2u_{2}-2=u_{2}\end{cases}}$

From this we can determine that ${\vec {u}}=\left\langle {\frac {1}{2}},0\right\rangle$

Exercise 1b: Variation of Parameters

$X'(t)={\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,X+{\begin{bmatrix}\mathrm {e} ^{t}\\t\end{bmatrix}}$

Solution to homogeneous system is $X_{h}(t)=c_{1}\,\mathrm {e} ^{t}\,{\begin{bmatrix}1\\1\end{bmatrix}}+c_{2}\,\mathrm {e} ^{-t}\,{\begin{bmatrix}1\\3\end{bmatrix}}$

Guess for particular solution:

${\begin{aligned}X_{p}(t)&=t\,\mathrm {e} ^{t}\,{\vec {u}}+t\,{\vec {v}}+{\vec {w}}+\mathrm {e} ^{t}\,{\vec {x}}\\&={\begin{bmatrix}u_{1}\,t\,\mathrm {e} ^{t}+t\,v_{1}+w_{1}+x_{1}\,\mathrm {e} ^{t}\\u_{2}\,t\,\mathrm {e} ^{-t}+t\,v_{2}+w_{2}+x_{2}\,\mathrm {e} ^{t}\end{bmatrix}}\end{aligned}}$

Find derivative and set equal to substituted RHS

${\begin{aligned}X_{p}'(t)&=\left(\mathrm {e} ^{t}+t\,\mathrm {e} ^{t}\right)\,{\vec {u}}+{\vec {v}}\\A\,X_{p}(t)+G(t)&=t\,\mathrm {e} ^{t}\,{\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,{\vec {u}}+t{\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,{\vec {v}}+{\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,{\vec {w}}+\mathrm {e} ^{t}\,{\begin{bmatrix}1\\0\end{bmatrix}}+t\,{\begin{bmatrix}0\\1\end{bmatrix}}\\\mathrm {e} ^{t}\,{\vec {u}}+{\vec {v}}&=\mathrm {e} ^{t}\,{\begin{bmatrix}2u_{1}-u_{2}\\3u_{1}-2u_{2}\end{bmatrix}}+t{\begin{bmatrix}2v_{1}-v_{2}\\3v_{1}-2v_{2}\end{bmatrix}}+{\begin{bmatrix}2w_{1}-w_{2}\\3w_{1}-2w_{2}\end{bmatrix}}+\mathrm {e} ^{t}\,{\begin{bmatrix}1\\0\end{bmatrix}}+t\,{\begin{bmatrix}0\\1\end{bmatrix}}\\&\vdots \\X_{p}(t)&=t\,\mathrm {e} ^{t}\,{\begin{bmatrix}{\frac {3}{2}}\\{\frac {3}{2}}\end{bmatrix}}+t\,{\begin{bmatrix}1\\2\end{bmatrix}}+{\begin{bmatrix}0\\-1\end{bmatrix}}+\mathrm {e} ^{t}\,{\begin{bmatrix}{\frac {1}{2}}\\0\end{bmatrix}}\end{aligned}}$

Therefore, the general solution is the sum $X(t)=X_{p}(t)+X_{h}(t)$:

$X(t)=t\,\mathrm {e} ^{t}\,{\begin{bmatrix}{\frac {3}{2}}\\{\frac {3}{2}}\end{bmatrix}}+t\,{\begin{bmatrix}1\\2\end{bmatrix}}+{\begin{bmatrix}0\\-1\end{bmatrix}}+\mathrm {e} ^{t}\,{\begin{bmatrix}{\frac {1}{2}}\\0\end{bmatrix}}+{\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,X+{\begin{bmatrix}\mathrm {e} ^{t}\\t\end{bmatrix}}$