Question

Solve the differential equation${\mathit{f}}^{\mathbf{\text{'}\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathit{f}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{+}\mathbf{13}\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathit{t}\mathbf{\right)}$and find all the real solutions of the differential equation

Short answer

The solution is$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}({\mathrm{C}}_1\cos 3\mathrm{t}+{\mathrm{C}}_2\sin 3\mathrm{t})+\frac{3}{40}\cos \left(\mathrm{t}\right)+\frac{1}{40}\sin \left(\mathrm{t}\right)$.

Step-by-step solution

Definition of characteristic polynomial

Consider the linear differential operator

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

The characteristic polynomial of is defined as

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

Determination of the solution

The characteristic polynomial of the operator as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}\text{'}+4\mathrm{f}\text{'}+13\mathrm{f}$

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^2+4\mathrm{\lambda }+13$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} \lambda =\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \lambda =\frac{-4\pm \sqrt{{\left(4\right)}^2-4\left(1\right)\left(13\right)}}{2\left(1\right)}\left\{\because a=1,,b=4,c=13\right\} \\ \lambda =\frac{-4\pm \sqrt{16-52}}{2} \\ \lambda =\frac{-4\pm \sqrt{-36}}{2} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{\lambda }=\frac{-4\pm 6\mathrm{i}}{2} \\ \mathrm{\lambda }=\frac{-4\pm 6\mathrm{i}}{2},\mathrm{\lambda }=\frac{-4-6\mathrm{i}}{2} \\ \mathrm{\lambda }=-2+3\mathrm{i},-2-3\mathrm{i} \end{gathered}$$

Therefore, the roots of the characteristic polynomials are 2+3i and 2-3i.

simplification of the solution

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

The exponential functions${\mathrm{e}}^{\left(-2+3\mathrm{i}\right)\mathrm{t}}$and${\mathrm{e}}^{\left(-2-3\mathrm{i}\right)\mathrm{t}}$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 exponential function${\mathrm{e}}^{\left(-2+3\mathrm{i}\right)\mathrm{t}}$and${\mathrm{e}}^{\left(-2-3\mathrm{i}\right)\mathrm{t}}$can be written as follows.

$$\begin{gathered} {\mathrm{e}}^{\left(-2+3\mathrm{i}\right)\mathrm{t}}={\mathrm{e}}^{-2\mathrm{t}}\left({\mathrm{c}}_1\cos 3\mathrm{t}+{\mathrm{c}}_2\sin 3\mathrm{t}\right) \\ {\mathrm{e}}^{\left(-2-3\mathrm{i}\right)\mathrm{t}}={\mathrm{e}}^{-2\mathrm{t}}\left({\mathrm{c}}_3\cos \left(-3\mathrm{t}\right)+{\mathrm{c}}_4\sin \left(-3\mathrm{t}\right)\right) \\ {\mathrm{e}}^{\left(-2-3\mathrm{i}\right)\mathrm{t}}={\mathrm{e}}^{-2\mathrm{t}}\left({\mathrm{c}}_2\cos 3\mathrm{t}-{\mathrm{c}}_4\sin 3\mathrm{t}\right) \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\left(-2+3\mathrm{i}\right)\mathrm{t}}+{\mathrm{e}}^{\left(-2-3\mathrm{i}\right)\mathrm{t}} \\ ={\mathrm{c}}_1{\mathrm{e}}^{-2\mathrm{t}}\cos 3\mathrm{t}+{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{t}}\sin 3\mathrm{t}+{\mathrm{c}}_3{\mathrm{e}}^{-2\mathrm{t}}\cos 3\mathrm{t}-{\mathrm{c}}_4{\mathrm{e}}^{-2\mathrm{t}}\sin 3\mathrm{t} \\ =\left({\mathrm{c}}_1+{\mathrm{c}}_3\right){\mathrm{e}}^{2\mathrm{t}}\cos 3\mathrm{t}+\left({\mathrm{c}}_2-{\mathrm{c}}_4\right){\mathrm{e}}^{2\mathrm{t}}\sin 3\mathrm{t} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\left({\mathrm{C}}_1\cos 3\mathrm{t}+{\mathrm{C}}_2\sin 3\mathrm{t}\right)\left\{\because {\mathrm{C}}_1={\mathrm{c}}_1+{\mathrm{c}}_3,{\mathrm{C}}_2={\mathrm{c}}_2-{\mathrm{c}}_4\right\} \end{gathered}$$

The complementary function as follows.

${\mathrm{f}}_{\mathrm{c}}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\left({\mathrm{C}}_1\cos 3\mathrm{t}+{\mathrm{C}}_2\sin 3\mathrm{t}\right)$

To find the particular solution

The particular solution to the differential equation$\mathrm{f}\text{'}\text{'}+4\mathrm{f}\text{'}+13\mathrm{f}=\cos \left(\mathrm{t}\right)$is of the form ${\mathrm{f}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right)$.

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

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

Similarly, differentiate the function$\mathrm{f}\text{'}=-\mathrm{Asint}+\mathrm{Bcost}$with respect to t 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${\mathrm{f}}^{\text{'}\text{'}}$and$-\mathrm{Asint}+\mathrm{Bcost}$for${\mathrm{f}}^{\text{'}}$and$\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right)$for f in$\mathrm{f}\text{'}\text{'}+4\mathrm{f}\text{'}+13\mathrm{f}$as follows.

$$\begin{gathered} {\mathrm{f}}^{\text{'}\text{'}}+4{\mathrm{f}}^{\text{'}}+13\mathrm{f}=-\mathrm{Acost}-\mathrm{Bsint}+4\left(-\mathrm{Asint}+\mathrm{Bcost}\right)+13\left(\mathrm{Acost}+\mathrm{Bsint}\right) \\ =\mathrm{Acost}-\mathrm{Bsint}-4\mathrm{Asint}+4\mathrm{Bscost}+13\mathrm{Acost}+13\mathrm{Bsint} \\ =12\mathrm{Acost}-12\mathrm{Bsint}-4\mathrm{Asint}+4\mathrm{Bcost} \\ {\mathrm{f}}^{\text{'}\text{'}}+4{\mathrm{f}}^{\text{'}}+13\mathrm{f}=\left(12\mathrm{A}+4\mathrm{b}\right)\mathrm{cost}+\left(-4\mathrm{A}+12\mathrm{B}\right)\mathrm{sint} \end{gathered}$$

Substitute the value $(12\mathrm{A}+4\mathrm{B})\mathrm{cost}+(-4\mathrm{A}+12\mathrm{B})\mathrm{sint}$for$\mathrm{f}\text{'}\text{'}+4\mathrm{f}\text{'}+13\mathrm{f}$in$\mathrm{f}\text{'}\text{'}+4\mathrm{f}\text{'}+13\mathrm{f}=\cos \left(\mathrm{t}\right)$as follows.

$$\begin{gathered} {\mathrm{f}}^{\text{'}\text{'}}+4{\mathrm{f}}^{\text{'}}+13\mathrm{f}=\cos \left(\mathrm{t}\right) \\ \left(12\mathrm{A}+4\mathrm{B}\right)\mathrm{cost}+\left(-4\mathrm{A}+12\mathrm{B}\right)\mathrm{sint}=\cos \left(\mathrm{t}\right) \end{gathered}$$

Compare the coefficients of and as follows.

$$\begin{gathered} 12\mathrm{A}+4\mathrm{B}=1 \\ -4\mathrm{A}+12\mathrm{B}=0 \end{gathered}$$

Solving the both the equations as follows.

$$\begin{gathered} 48\mathrm{A}+16\mathrm{B}=4 \\ \frac{-48\mathrm{A}+144\mathrm{B}=0}{160\mathrm{B}=4} \\ \mathrm{B}=\frac{4}{160} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{B}=\frac{4}{160} \\ \mathrm{B}=\frac{1}{40} \end{gathered}$$

Substitute the value $\frac{1}{40}$for B in 12A+4B=1 as follows.

$$\begin{gathered} 12\mathrm{A}+4\mathrm{B}=1 \\ 12\mathrm{A}+4\frac{1}{40}=1 \\ 12\mathrm{A}+\frac{1}{10}=1 \\ 12\mathrm{A}=1-\frac{1}{10} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} 12A=\frac{10-1}{10} \\ 12A=\frac{9}{10} \\ A=\frac{9}{120} \\ A=\frac{3}{40} \end{gathered}$$

Substitute the value$\frac{3}{40}$ for A and$\frac{1}{40}$ 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.

$$\begin{gathered} {\mathrm{f}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)+\mathrm{Bsin}\left(\mathrm{t}\right) \\ {\mathrm{f}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{3}{40}\cos \left(\mathrm{t}\right)+\frac{1}{40}\sin \left(\mathrm{t}\right) \end{gathered}$$

Therefore, the general solution as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{f}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{f}}_{\mathrm{p}}\left(\mathrm{t}\right) \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\left({\mathrm{C}}_1\cos 3\mathrm{t}+{\mathrm{C}}_2\sin 3\mathrm{t}\right)+\frac{3}{40}\cos \left(\mathrm{t}\right)+\frac{1}{40}\sin \left(\mathrm{t}\right) \end{gathered}$$

Thus, the general solution of the differential equation$\mathrm{f}\text{'}\text{'}+4\mathrm{f}\text{'}+13\mathrm{f}=\cos \left(\mathrm{t}\right)$ is$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\left({\mathrm{C}}_1\cos 3\mathrm{t}+{\mathrm{C}}_2\sin 3\mathrm{t}\right)+\frac{3}{40}\cos \left(\mathrm{t}\right)+\frac{1}{40}\sin \left(\mathrm{t}\right)$