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Found in: Page 442

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

Solve the differential equation ${\mathbf{f}}{\mathbf{\text{'}\text{'}\text{'}}}\left(t\right){\mathbf{-}}{\mathbf{f}}{\mathbf{\text{'}\text{'}}}\left(t\right){\mathbf{-}}{\mathbf{4}}{\mathbf{f}}{\mathbf{\text{'}}}\left(t\right){\mathbf{+}}{\mathbf{4}}{\mathbf{f}}\left(t\right){\mathbf{=}}{\mathbf{0}}$and find all the real solution of the differential equation.

The solution is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_{1}{\mathrm{e}}^{\mathrm{t}}+{\mathrm{C}}_{2}{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{C}}_{3}{\mathrm{e}}^{-2\mathrm{t}}$.

See the step by step solution

Step1: Definition of characteristic polynomial

Consider the linear differential operator

$\mathrm{T}\left(\mathrm{f}\right)={\mathrm{f}}^{\left(\mathrm{n}\right)}+{\mathrm{a}}_{\mathrm{n}-1}{\mathrm{f}}^{\left(\mathrm{n}-1\right)}+...+{\mathrm{a}}_{1}\mathrm{f}\text{'}+{\mathrm{a}}_{0}\mathrm{f}$.

The characteristic polynomial of T is defined as

${\mathrm{p}}_{\mathrm{T}}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^{\mathrm{n}}+{\mathrm{a}}_{\mathrm{n}-1}+...+{\mathrm{a}}_{1}\mathrm{\lambda }+{\mathrm{a}}_{0}$.

Step2: Determination of the solution

The characteristic polynomial of the operator as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}\text{'}\text{'}‐\mathrm{f}\text{'}\text{'}‐4\mathrm{f}\text{'}+4\mathrm{f}$

Then the characteristic polynomial is as follows.

role="math" localid="1660798972108" ${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^{3}‐{\mathrm{\lambda }}^{2}‐4\mathrm{\lambda }+4$

Solve the characteristic polynomial and find the roots as follows.

${\lambda }^{3}‐{\lambda }^{2}‐4\lambda +4=0\phantom{\rule{0ex}{0ex}}{\lambda }^{2}\left(\lambda ‐1\right)‐4\left(\lambda ‐1\right)=0\phantom{\rule{0ex}{0ex}}\left({\lambda }^{2}‐4\right)\left(\lambda ‐1\right)=0\phantom{\rule{0ex}{0ex}}\left(\lambda +2\right)\left(\lambda ‐2\right)\left(\lambda ‐1\right)=0$

Simplify further as follows.

$\lambda ‐1=0\phantom{\rule{0ex}{0ex}}\lambda =1\phantom{\rule{0ex}{0ex}}\lambda +2=0\phantom{\rule{0ex}{0ex}}\lambda =‐2$

Simplify further as follows.

$\lambda ‐2=0\phantom{\rule{0ex}{0ex}}\lambda =2$

Therefore, the roots of the characteristic polynomials are 1,2 and $‐2$.

Step3: Conclusion of the solution

Since, the roots of the characteristic equation is different real numbers.

The exponential functions ${\mathrm{e}}^{\mathrm{t}},{\mathrm{e}}^{2\mathrm{t}}$ and ${\mathrm{e}}^{-2\mathrm{t}}$ form a basis of the kernel of T .

Hence, they form a basis of the solution space of the homogenous differential equation is $\mathrm{T}\left(\mathrm{f}\right)=0$.

Thus, the general solution of the differential equation $\mathrm{f}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)‐\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)‐4\mathrm{f}\text{'}\left(\mathrm{t}\right)+4\mathrm{f}\text{'}\left(\mathrm{t}\right)=0$ is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_{1}{\mathrm{e}}^{\mathrm{t}}+{\mathrm{C}}_{2}{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{C}}_{3}{\mathrm{e}}^{-2\mathrm{t}}$.