Question

Solve the differential equation$\mathbf{f}\mathbf{\text{'}}\mathbf{\text{'}}\left(t\right)\mathbf{+}\mathbf{f}\mathbf{\text{'}}\left(t\right)\mathbf{-}\mathbf{12}\mathbf{f}\left(t\right)\mathbf{=}\mathbf{0}$and find solution of the differential equation.

Short answer

The solution is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}$.

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 T 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.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}\text{'}+\mathrm{f}\text{'}-12$

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^2+\mathrm{\lambda }‐12$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} \lambda =\frac{‐b\pm \sqrt{b^2‐4ac}}{2a} \\ \lambda =\frac{‐1\pm \sqrt{{\left(1\right)}^2‐4\left(1\right)\left(‐12\right)}}{2\left(1\right)}\left\{\because a=1,,b=1,c=‐12\right\} \\ \lambda =\frac{‐1\pm \sqrt{1+48}}{2} \\ \lambda =\frac{‐1\pm \sqrt{49}}{2} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \lambda =\frac{‐1\pm 7}{2} \\ \lambda =\frac{‐1\pm 7}{2},\lambda =\frac{‐1‐7}{2} \\ \lambda =\frac{6}{2},\lambda =\frac{‐8}{2} \\ \lambda =3,‐4 \end{gathered}$$

Therefore, the roots of the characteristic polynomials are 3 and $‐4$.

Conclusion of the solution

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

The exponential functions ${\mathrm{e}}^{3\mathrm{t}}$and${\mathrm{e}}^{-4\mathrm{t}}$ form a basis of the kernel of T .

Hence, they form a basis of the solution space of the homogenous differential equation is $\mathrm{T}\left(\mathrm{f}\right)=0$.

Thus, the general solution of the differential equation$\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)+\mathrm{f}\text{'}\left(\mathrm{t}\right)‐12=0$ is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}$.