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Expert-verified Found in: Page 442 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Solve the differential equation $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{x}}{\mathbf{=}}{\mathbf{cos}}\left(t\right)$ and find all the real solutions of the differential equation.

The solution is $\mathrm{x}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\mathrm{cos}\left(\sqrt{2}\mathrm{t}\right)+{\mathrm{c}}_{2}{\mathrm{e}}^{\mathrm{t}}\mathrm{sin}\left(\sqrt{2}\mathrm{t}\right)+\mathrm{cos}\left(\mathrm{t}\right)$ .

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.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{x}"+2\mathrm{x}$

Then the characteristic polynomial is as follows.

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

Solve the characteristic polynomial and find the roots as follows.

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

Therefore, the roots of the characteristic polynomials are $\sqrt{2\mathrm{i}}$ and $-\sqrt{2\mathrm{i}}$ .

## Step3: Simplification of the solution

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

The exponential functions ${\mathrm{e}}^{\sqrt{2\mathrm{it}}}$ and ${\mathrm{e}}^{-\sqrt{2\mathrm{it}}}$ 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 .

The complementary function as follows.

${\mathrm{f}}_{\mathrm{c}}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\mathrm{cos}\left(\sqrt{2\mathrm{t}}\right)+{\mathrm{c}}_{2}{\mathrm{e}}^{\mathrm{t}}\mathrm{sin}\left(\sqrt{2\mathrm{t}}\right)$

## Step4: To find the particular solution

The particular solution to the differential equation $\mathrm{x}"+2\mathrm{x}+=\mathrm{cos}\left(\mathrm{t}\right)$ is of the form role="math" localid="1660806105789" ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right)$ .

Differentiate the function ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right)$ with respect to t as follows.

${x}_{p}\left(t\right)=A\mathrm{cos}\left(t\right)+B\mathrm{sin}\left(t\right)\phantom{\rule{0ex}{0ex}}x\text{'}=-A\mathrm{sin}+B\mathrm{cos}t$

Similarly, differentiate the function $x\text{'}=-A\mathrm{sin}t+B\mathrm{cos}t$ with respect to t as follows.

$x\text{'}=-A\mathrm{sin}t+B\mathrm{cos}t\phantom{\rule{0ex}{0ex}}x"=-A\mathrm{cos}t-B\mathrm{sin}t$

Substitute the values $-A\mathrm{cos}t-B\mathrm{sin}t$ for x" and $A\mathrm{cos}\left(t\right)+B\mathrm{sin}\left(t\right)$ for x in x"+2x as follows.

$\mathrm{x}"+2\mathrm{x}=-\mathrm{Acost}-\mathrm{Bsint}+2\left(\mathrm{Acost}+\mathrm{Bsint}\right)\phantom{\rule{0ex}{0ex}}=-\mathrm{Acost}-\mathrm{Bsint}+2\mathrm{Acost}+2\mathrm{Bsint}\phantom{\rule{0ex}{0ex}}=\mathrm{Acost}+\mathrm{Bsint}\phantom{\rule{0ex}{0ex}}\mathrm{x}"+2\mathrm{x}=\mathrm{Acost}+\mathrm{Bsint}$

Substitute the value $A\mathrm{cos}t+B\mathrm{sin}t$ for $\mathrm{x}"+2\mathrm{x}$ in x"+2x = cos(t) as follows.

$\mathrm{x}"+2\mathrm{x}=\mathrm{cos}\left(\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Acost}+\mathrm{Bsint}=\mathrm{cos}\left(\mathrm{t}\right)$

Compare the coefficients of Cost and sint as follows.

A = 1

B = 0

Substitute the value 1 for A and 0 for B in role="math" localid="1660806778527" ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right)$ as follows.

${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=1.\mathrm{cos}\left(\mathrm{t}\right)+0.\mathrm{sin}\left(\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{cos}\left(\mathrm{t}\right)$

Therefore, the general solution as follows.

$\mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}\mathrm{x}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\mathrm{cos}\left(\sqrt{2}\mathrm{t}\right)+{\mathrm{c}}_{2}{\mathrm{e}}^{\mathrm{t}}\mathrm{sin}\left(\sqrt{2}\mathrm{t}\right)+\mathrm{cos}\left(\mathrm{t}\right)$

Thus, the general solution of the differential equation $\mathrm{x}"+2\mathrm{x}=\mathrm{cos}\left(\mathrm{t}\right)$ is $\mathrm{x}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\mathrm{cos}\left(\sqrt{2}\mathrm{t}\right)+{\mathrm{c}}_{2}{\mathrm{e}}^{\mathrm{t}}\mathrm{sin}\left(\sqrt{2\mathrm{t}}\right)+\mathrm{cos}\left(\mathrm{t}\right)$ . ### Want to see more solutions like these? 