91Ó°ÊÓ

Consider the linear differential operator

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

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.

$T\left(f\right)=f"\left(t\right)+4f\left(t\right)$

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{λ}\right)={\mathrm{λ}}^2+4$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} \mathrm{λ}=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a} \\ \mathrm{λ}=\frac{-0Â\pm \sqrt{{\left(0\right)}^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}\left\{âˆ\mu a=1,,b=0,c=4\right\} \\ \mathrm{λ}=\frac{-0Â\pm \sqrt{-16}}{2} \\ \mathrm{λ}=\frac{4i}{2} \\ \mathrm{λ}=Â\pm 2i \end{gathered}$$

$\begin{array}{l}\mathrm{λ}=\frac{4i}{2}\\ \mathrm{λ}=Â\pm 2i\end{array}$

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

Since, the roots of the characteristic equation are different complex numbers.

The exponential functions$e^{2t}$and $e^{-2t}$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 exponential function $e^{2t}$and$e^{-2t}$ can be written as follows.

Simplify further as follows.

role="math" localid="1661143288908" $\begin{array}{l}{e}^{2it}=C_1\cos \left(2t\right)+C_2,\sin (2t\\ e^{-2it}=C_{3}cos(-2t)+C_4,\sin \left(22t\right\}\\ e^{-2it}=C_3\cos \left(2t\right)-C_4\sin \left(2t\right)\end{array}$

The complementary function as follows.

$$\begin{gathered} f\left(t\right)=e^{2it}+e^{-2it} \\ =C_1\cos \left(2t\right)+C_2\sin \left(2t\right)+C_3\cos (-2t)+C_4\sin (-2t) \\ =C_1\cos \left(2t\right)+C_2\sin \left(2t\right)+C_3\cos \left(2t\right)-C_4\sin \left(2t\right) \\ =\left(C_1+C_2\right)\sin \left(2t\right)+\left(C_3-C_4\right)\sin \left(2t\right) \end{gathered}$$

$$\begin{gathered} âˆ\mu C_1=C_1+C_2 \\ âˆ\mu C_2=C_3-C_4 \\ {f}_c\left(t\right)=C_1\sin \left(2t\right)+C_2\sin \left(2t\right) \end{gathered}$$

The particular solution to the differential equation $f"\left(t\right)+4f\left(t\right)=\sin t$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 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$f\text{'}=-A\cos \left(t\right)+B\sin \left(t\right)$ with respect to as follows.

role="math" localid="1661141606101" $$\begin{gathered} f\text{'}=-A\cos \left(t\right)+B\cos \left(t\right) \\ f"=-A\cos \left(t\right)-B\sin \left(t\right) \end{gathered}$$

Substitute the values $-\mathrm{Acos}\left(\mathrm{t}\right)-\mathrm{Bsin}\left(\mathrm{t}\right)$for $f"$and$-A\sin t+B\cos t$for f' and role="math" localid="1661141624863" $A\cos t+B\sin t$ for in as follows.

$$\begin{gathered} f"+4f=-A\cos t+B\sin t+4\left(A\cos t+B\sin t\right) \\ =-A\cos t+B\sin t+4A\cos t+4B\sin t \\ =3A\cos t+3B\sin t \\ f"+4f=\left(3A\cos t+3B\sin t\right) \end{gathered}$$

Substitute the value (3A+3B) for $f"+4f$in $f"\left(t\right)+4f\left(t\right)=\sin t$as follows.

$$\begin{gathered} f"+4f=\sin \left(t\right) \\ 3A\cos t+3B\sin t=\sin \left(t\right) \end{gathered}$$

Compare the coefficients of and as follows.

3A=0

3B=1

Simplify further as follows.

$\begin{array}{l}3B=1\\ B=\frac{1}{3}\end{array}$

A=0

Substitute the value 0 for A and $\frac{1}{3}$for B in$f_p\left(t\right)=A\cos \left(t\right)+B\sin \left(t\right)$ as follows.

role="math" localid="1661141157207" $\begin{array}{l}{f}_p\left(t\right)=A\cos \left(t\right)+B\sin \left(t\right)\\ f_p\left(t\right)=\left(0\right)\cos \left(t\right)+\frac{1}{3}\sin \left(t\right)\end{array}$

Therefore, the general solution as follows.

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

Thus, the general solution of the differential equationrole="math" localid="1661140980597" $f^{"}\left(t\right)+4f\left(t\right)=\sin t$is

$f\left(t\right)=C_1\sin \left(2t\right)+C_2\sin \left(2t\right)+\frac{1}{3}\sin \left(t\right)$.

Consider the equation $f"\left(t\right)=C_1\sin \left(2t\right)=\sin t$with $f\left(t\right)=C_1\sin \left(2t\right)+C_2\sin \left(2t\right)+\frac{1}{3}\sin \left(t\right)$

role="math" localid="1661142904593" $\begin{array}{l}f\left(t\right)=f_c\left(t\right)+f_p\left(t\right)\\ f\left(t\right)=C_1\sin \left(2t\right)+C_2\sin \left(2t\right)+\frac{1}{3}\sin \left(t\right)\end{array}$

Here$C_1=\frac{-1}{2}and{C}_2=\frac{1}{3}$

$$\begin{gathered} f\left(t\right)=C_1\sin \left(2t\right)+C_2\sin \left(2t\right)+\frac{1}{3}\sin \left(t\right) \\ f\left(t\right)=\left(C_1+C_2\right)\sin \left(2t\right)+\frac{1}{3}\sin \left(t\right) \\ f\left(t\right)=\left(\frac{-1}{2}+\frac{1}{3}\right)\sin \left(2t\right)+\frac{1}{3}\sin \left(t\right) \end{gathered}$$

Simplify further,

$\begin{array}{l}f\left(t\right)=-\frac{1}{6}\sin \left(2t\right)+\frac{1}{3}\sin \left(t\right)\\ f\left(t\right)=\frac{1}{3}\sin \left(t\right)-\frac{1}{6}\sin \left(2t\right)\end{array}$

Hence the solution is $f\left(t\right)=\frac{1}{3}\sin t-\frac{1}{6}\sin \left(2t\right)$.