91Ó°ÊÓ

Consider the linear differential operator

$\mathrm{T}\left(\mathrm{f}\right)={\mathrm{f}}^{\left(\mathrm{n}\right)}+{\mathrm{a}}_{\mathrm{n}-1}+...+{\mathrm{a}}_1\mathrm{f}\text{'}+{\mathrm{a}}_0\mathrm{f}$.

The characteristic polynomial of T 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.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}+3\mathrm{f}\text{'}+2\mathrm{f}$

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{λ}\right)={\mathrm{λ}}^2+3\mathrm{λ}+2$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} {\mathrm{λ}}^2+3\mathrm{λ}+2=0 \\ {\mathrm{λ}}^2+\mathrm{λ}+2\mathrm{λ}+2=0 \\ \mathrm{λ}\left(\mathrm{λ}+1\right)+2\left(\mathrm{λ}+1\right)=0 \\ \left(\mathrm{λ}+2\right)\left(\mathrm{λ}+1\right)=0 \end{gathered}$$

Simplify further as follows.

$\begin{array}{l}\mathrm{λ}+1=0\\ \mathrm{λ}=-1\\ \mathrm{λ}+2=0\\ \mathrm{λ}=-2\end{array}$

Therefore, the roots of the characteristic polynomials are -1 and -2 .

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

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

The complementary function as follows.

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

The particular solution to the differential equation$\mathrm{f}"+3\mathrm{f}\text{'}+2\mathrm{f}=\cos \left(\mathrm{t}\right)$is of the formlocalid="1661140531244" ${\mathrm{f}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right)$.

Differentiate the functionlocalid="1661140524869" ${\mathrm{f}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right)$with respect to as follows.

localid="1661140556797" $$\begin{gathered} {\mathrm{f}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right) \\ \mathrm{f}\text{'}=-\mathrm{Asint}+\mathrm{Bcost} \end{gathered}$$

Similarly, differentiate the function$\mathrm{f}\text{'}=-\mathrm{Asint}+\mathrm{Bcost}$
with respect to as follows.

$$\begin{gathered} \mathrm{f}\text{'}=-\mathrm{Asint}+\mathrm{Bcost} \\ \mathrm{f}\text{'}=-\mathrm{Acost}+\mathrm{Bsint} \end{gathered}$$

Similarly, differentiate the function$\mathrm{f}\text{'}=-\mathrm{Asint}+\mathrm{Bcost}$with respect to as follows.

$$\begin{gathered} \mathrm{f}\text{'}=-\mathrm{Asint}+\mathrm{Bcost} \\ \mathrm{f}\text{'}\text{'}=-\mathrm{Acost}-\mathrm{Bsint} \end{gathered}$$

Substitute the values$-\mathrm{Acost}-\mathrm{Bsint}$for f" and$-\mathrm{Asint}+\mathrm{Bcost}$for f' and$\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right)$for in f' + 3f' + 2f as follows.

$$\begin{gathered} \mathrm{f}"+3\mathrm{f}\text{'}+2\mathrm{f}=-\mathrm{Acost}-\mathrm{Bsint}+3(-\mathrm{Asint}+\mathrm{Bcost})+2(\mathrm{Acost}+\mathrm{Bsint}) \\ =-\mathrm{Acost}-\mathrm{Bsint}-3\mathrm{Asint}+3\mathrm{Bcost}+2\mathrm{Acost}+2\mathrm{Bsint} \\ =\mathrm{Acost}-3\mathrm{Asint}+3\mathrm{Bcost}+\mathrm{Bsint} \\ \mathrm{f}"+2\mathrm{f}\text{'}+\mathrm{f}=\left(\mathrm{A}+3\mathrm{B}\right)\mathrm{cost}+(-3\mathrm{A}+\mathrm{B})\mathrm{sint} \end{gathered}$$

Substitute the value $\left(\mathrm{A}+3\mathrm{B}\right)\mathrm{cost}+\left(-3\mathrm{A}+\mathrm{B}\right)\mathrm{sint}$for f' + 3f' + 2f in f" + 3f' + 2f = cos (t) as follows.

$$\begin{gathered} \mathrm{f}"+3\mathrm{f}\text{'}+2\mathrm{f}=\cos \left(\mathrm{t}\right) \\ \left(\mathrm{A}+3\mathrm{B}\right)\mathrm{cost}+\left(-3\mathrm{A}+\mathrm{B}\right)\mathrm{sint}=\mathrm{cost}\left(\mathrm{t}\right) \end{gathered}$$

Compare the coefficients of and as follows.

A + 3B =1

-3A + B =0

Solving the both the equations as follows.

$$\begin{gathered} 3\mathrm{A}+9\mathrm{B}=3 \\ \frac{-3\mathrm{A}+\mathrm{B}=0}{10\mathrm{B}=3} \\ \mathrm{B}=\frac{3}{10} \end{gathered}$$

Substitute the value$\frac{3}{10}$for B in A + 3B = 1 as follows.

$$\begin{gathered} \mathrm{A}+3\mathrm{B}=1 \\ \mathrm{A}+3\frac{3}{10}=1 \\ \mathrm{A}+\frac{9}{10}=1 \\ \mathrm{A}=1-\frac{9}{10} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{A}=\frac{10-9}{10} \\ \mathrm{A}=\frac{1}{10} \end{gathered}$$

Substitute the value $\frac{1}{10}$for A and $\frac{3}{10}$for B in${\mathrm{f}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right)$as follows.

$$\begin{gathered} {\mathrm{f}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right) \\ {\mathrm{f}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{1}{10}\cos \left(\mathrm{t}\right)+\frac{3}{10}\sin \left(\mathrm{t}\right) \end{gathered}$$

Therefore, the general solution as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{f}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{f}}_{\mathrm{p}}\left(\mathrm{t}\right) \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{C}}_2{\mathrm{e}}^{-2\mathrm{t}}+\frac{1}{10}\cos \left(\mathrm{t}\right)+\frac{3}{10}\sin \left(\mathrm{t}\right) \end{gathered}$$

Thus, the general solution of the differential equation$\mathrm{f}"+3\mathrm{f}\text{'}+2\mathrm{f}=\cos \left(\mathrm{t}\right)$ is$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{C}}_2{\mathrm{e}}^{-2\mathrm{t}}+\frac{1}{10}\cos \left(\mathrm{t}\right)+\frac{3}{10}\sin \left(\mathrm{t}\right)$ .