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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 ${\mathbit{f}}{\mathbf{"}}\left(t\right){\mathbf{+}}{\mathbf{2}}{\mathbit{f}}{\mathbf{\text{'}}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{+}}{\mathbit{f}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}$ and find all the real solutions of the differential equation.

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

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.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}"-2\mathrm{f}\text{'}+\mathrm{f}$

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^{2}-2\mathrm{\lambda }+1$

Solve the characteristic polynomial and find the roots as follows.

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

Simplify further as follows.

$\mathrm{\lambda }+1=0\phantom{\rule{0ex}{0ex}}\mathrm{\lambda }=-1\phantom{\rule{0ex}{0ex}}\mathrm{\lambda }+1=0\phantom{\rule{0ex}{0ex}}\mathrm{\lambda }=-1$

Therefore, the roots of the characteristic polynomials are -1 and -1 .

## Step3: Conclusion of the solution

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

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

Hence, they form a basis of the solution space of the homogenous differential equation is T(f) = 0 .

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