91Ó°ÊÓ

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

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{λ}\right)={\mathrm{λ}}^{\mathrm{n}}+{\mathrm{a}}_{\mathrm{n}-1}\mathrm{λ}+...+{\mathrm{a}}_1\mathrm{λ}+{\mathrm{a}}_0$.

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.

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

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} {\mathrm{λ}}^2+2=0 \\ {\mathrm{λ}}^2=-2 \\ \mathrm{λ}=Â\pm \sqrt{2\mathrm{i}} \end{gathered}$$

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

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.

${\mathrm{f}}_{\mathrm{c}}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(\sqrt{2\mathrm{t}}\right)+{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{t}}\sin \left(\sqrt{2\mathrm{t}}\right)$

The particular solution to the differential equation$\mathrm{x}"+2\mathrm{x}+=\cos \left(\mathrm{t}\right)$is of the formrole="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.

$$\begin{gathered} x_p\left(t\right)=A\cos \left(t\right)+B\sin \left(t\right) \\ x\text{'}=-A\sin +B\cos t \end{gathered}$$

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

$$\begin{gathered} x\text{'}=-A\sin t+B\cos t \\ x"=-A\cos t-B\sin t \end{gathered}$$

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

$$\begin{gathered} \mathrm{x}"+2\mathrm{x}=-\mathrm{Acost}-\mathrm{Bsint}+2\left(\mathrm{Acost}+\mathrm{Bsint}\right) \\ =-\mathrm{Acost}-\mathrm{Bsint}+2\mathrm{Acost}+2\mathrm{Bsint} \\ =\mathrm{Acost}+\mathrm{Bsint} \\ \mathrm{x}"+2\mathrm{x}=\mathrm{Acost}+\mathrm{Bsint} \end{gathered}$$

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

$$\begin{gathered} \mathrm{x}"+2\mathrm{x}=\cos \left(\mathrm{t}\right) \\ \mathrm{Acost}+\mathrm{Bsint}=\cos \left(\mathrm{t}\right) \end{gathered}$$

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.

$$\begin{gathered} {\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right) \\ {\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=1.\cos \left(\mathrm{t}\right)+0.\sin \left(\mathrm{t}\right) \\ {\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=\cos \left(\mathrm{t}\right) \end{gathered}$$

Therefore, the general solution as follows.

$$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right) \\ \mathrm{x}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(\sqrt{2}\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{t}}\sin \left(\sqrt{2}\mathrm{t}\right)+\cos \left(\mathrm{t}\right) \end{gathered}$$

Thus, the general solution of the differential equation$\mathrm{x}"+2\mathrm{x}=\cos \left(\mathrm{t}\right)$ is$\mathrm{x}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(\sqrt{2}\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{t}}\sin \left(\sqrt{2\mathrm{t}}\right)+\cos \left(\mathrm{t}\right)$ .