91Ó°ÊÓ

Consider the linear differential operator

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

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.

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

Then the characteristic polynomial is as follows.

${\mathrm{p}}_{\mathrm{T}}\left(\mathrm{λ}\right)={\mathrm{λ}}^2+2\mathrm{λ}+1$

Solve the characteristic polynomial and find the roots as follows.

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

Simplify further as follows.

$$\begin{gathered} \mathrm{λ}+1=0 \\ \mathrm{λ}=-1 \\ \mathrm{λ}+1=0 \\ \mathrm{λ}=-1 \end{gathered}$$

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

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

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

The complementary function as follows.

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

The particular solution to the differential equation $f\text{'}+2f\text{'}+f=\sin \left(t\right)$is of the form $f_p\left(t\right)=A\cos \left(t\right)+B\sin \left(t\right)$.

Differentiate the function $f_p\left(t\right)=A\cos \left(t\right)+B\sin \left(t\right)$with respect to t as follows.

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

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

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

Substitute the values -Acost-Bsint for f" and -Asint+Bcost for f' and Acos(t)+Bsin(t) for f in f"+2f'+f as follows.

$$\begin{gathered} f"+2f\text{'}+f=-Acost-B\sin t+2\left(Acost+B\sin t\right)+\left(A\cos t+B\sin t\right) \\ =-Acost-B\sin t-2Acost+2B\sin t+A\cos t+B\sin t \\ =-2A\sin t+2B\cos t \\ f"+2f\text{'}+f=-2A\sin t+2B\cos t \end{gathered}$$

Substitute the value -2Asint+2Bcost for f"+2f'+f in f"+2f'+f=sin(t) as follows.

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

Compare the coefficients of Cost and sint as follows.

$\begin{array}{l}-2A=1\\ 2B=0\end{array}$

Solving the both the equations as follows.

$\begin{array}{l}2B=0\\ B=0\end{array}$

Simplify the other equation as follows.

$\begin{array}{l}-2A=1\\ A=\frac{1}{2}\end{array}$

Substitute the value $\frac{-1}{2}$for A and 0 for B in $f_p\left(t\right)=A\cos \left(t\right)+B\sin \left(t\right)$as follows.

$$\begin{gathered} f_p\left(t\right)=A\cos \left(t\right)+B\sin \left(t\right) \\ {f}_p\left(t\right)=\frac{-1}{2}\cos \left(t\right)+0\sin \left(t\right) \\ {f}_p\left(t\right)=\frac{-1}{2}\cos \left(t\right) \end{gathered}$$

Therefore, the general solution as follows.

$\begin{array}{l}f\left(t\right)=f_c\left(t\right)+f_p\left(t\right)\\ f\left(t\right)=e^{-t}+(C_1+C_{2}t)-\frac{1}{2}\cos \left(t\right)\end{array}$

Thus, the general solution of the differential equation f"+2f'+f=sin(t) is $f\left(t\right)=e^{-t}\left(C_1+C_{2}t\right)-\frac{1}{2}\cos \left(t\right)$.