Question

Solve the differential equation $\mathbf{f}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$and find all the real solutions of the differential equation.

Short answer

The solution is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1\mathrm{cost}+{\mathrm{C}}_2\mathrm{sint}$.

Step-by-step solution

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

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

Determination of the solution

The characteristic polynomial of the operator as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}\text{'}+1$

Then the characteristic polynomial is as follows.

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

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} {\mathrm{\lambda }}^2-1=0 \\ {\mathrm{\lambda }}^2=-1 \\ \mathrm{\lambda }=\mathrm{i}\sqrt{1} \\ \mathrm{\lambda }=\pm \mathrm{i} \end{gathered}$$

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

Conclusion of the solution

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

The exponential functions $e^{0+\mathrm{it}}$and$e^{0-\mathrm{it}}$form a basis of the kernel of .

Hence, they form a basis of the solution space of the homogenous differential equation is T (f) =0.

The exponential functions$e^{0+\mathrm{it}}$and $e^{0-\mathrm{it}}$can be written as follows.

localid="1660815725795" $$\begin{gathered} e^{0+\mathrm{it}}={\mathrm{e}}^0\left({\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}\right) \\ ={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint} \\ {\mathrm{e}}^{0-\mathrm{it}}={\mathrm{e}}^0\left({\mathrm{c}}_3\cos \left(-\right)+{\mathrm{c}}_4\sin \left(-\mathrm{t}\right)\right) \\ ={\mathrm{c}}_3\mathrm{cost}+{\mathrm{c}}_4\mathrm{sint}\left\{\because \cos \left(-\mathrm{\theta }\right)\mathrm{os\theta },\sin \left(\mathrm{\theta }\right)=-\mathrm{sin\theta }\right\} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{it}}+{\mathrm{e}}^{-\mathrm{it}} \\ ={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}+{\mathrm{c}}_3\mathrm{cost}+{\mathrm{c}}_4\mathrm{sint} \\ =\left({\mathrm{c}}_1+{\mathrm{c}}_3\right)\mathrm{cost}+\left({\mathrm{c}}_2-{\mathrm{c}}_4\right)\mathrm{sint} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}\left\{\because {\mathrm{C}}_1={\mathrm{c}}_1+{\mathrm{c}}_3,{\mathrm{C}}_2={\mathrm{c}}_2-{\mathrm{c}}_4\right\} \end{gathered}$$

Thus, the general solution of the differential equation $\mathrm{f}"\left(\mathrm{t}\right)+\mathrm{f}\left(\mathrm{t}\right)=0\mathrm{is}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1\mathrm{cost}+{\mathrm{C}}_2\mathrm{sint}$