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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 ${\mathbit{f}}{\mathbf{"}}\left(t\right){\mathbf{=}}{\mathbf{0}}$ and find all the real solutions of the differential equation.

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

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}"$

Then the characteristic polynomial is as follows.

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

Solve the characteristic polynomial and find the roots as follows.

$\begin{array}{l}{\mathrm{\lambda }}^{2}=0\\ \mathrm{\lambda }=0\\ \mathrm{\lambda }=0,\mathrm{\lambda }=0\end{array}$

Therefore, the roots of the characteristic polynomials are 0 and 0.

## Step3: Conclusion of the solution

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

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