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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 initial value problem in ${\mathbf{f}}{\mathbf{\text{'}\text{'}}}\left(t\right){\mathbf{‐}}{\mathbf{9}}{\mathbf{f}}\left(t\right){\mathbf{=}}{\mathbf{0}}{\mathbf{;}}{\mathbf{f}}\left(0\right){\mathbf{=}}{\mathbf{0}}{\mathbf{,}}{\mathbf{f}}{\mathbf{\text{'}}}\left(0\right){\mathbf{=}}{\mathbf{1}}$

The solution is $f\left(t\right)=\mathrm{cos}\left(3\mathrm{t}\right)$.

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 is defined as

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

## Step2: Determination of the solution

The characteristic polynomial of the operator as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}-9$

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^{2}-9$

Solve the characteristic polynomial and find the roots as follows.

${\mathrm{\lambda }}^{2}-9=0\phantom{\rule{0ex}{0ex}}{\mathrm{\lambda }}^{2}=9\phantom{\rule{0ex}{0ex}}\mathrm{\lambda }=\sqrt{9}\phantom{\rule{0ex}{0ex}}\mathrm{\lambda }=±3\phantom{\rule{0ex}{0ex}}$

Therefore, the roots of the characteristic polynomials are 3 and -3.

## Step3: Explanation of the solution

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

The exponential functions ${\mathrm{e}}^{3\mathrm{t}}\mathrm{and}{\mathrm{e}}^{-3\mathrm{t}}$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 $\mathrm{f}\text{'}\left(\mathrm{t}\right)-9\mathrm{f}\text{'}\left(\mathrm{t}\right)=0\mathrm{is}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-3\mathrm{t}}$ .

## Step 4: Explanation of the solution with initial value problem

Consider the general solution of the $\mathrm{f}\text{'}"\left(\mathrm{t}\right)-9\mathrm{f}"\left(\mathrm{t}\right)=0$ with the initial value problem $f\left(0\right)=0,f\text{'}\left(0\right)=1$ is as follows

role="math" localid="1660809514458" $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-3\mathrm{t}}\phantom{\rule{0ex}{0ex}}\mathrm{f}\left(\mathrm{t}\right)=0\left({\mathrm{e}}^{3\mathrm{t}}\right)+1\left({\mathrm{e}}^{-3\mathrm{t}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{f}\left(\mathrm{t}\right)=0+{\mathrm{e}}^{-3\mathrm{t}}\phantom{\rule{0ex}{0ex}}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-3\mathrm{t}}$

Simplify further.

$=\mathrm{cos}\left(3t\right)$

Hence the solution is $f\left(t\right)=\mathrm{cos}\left(3t\right)$.