Question

Solve the differential equation$\mathit{f}\mathbf{"}\left(t\right)\mathbf{=}\mathbf{0}$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{C}}_2\mathrm{t}$$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1+{\mathrm{C}}_2\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 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.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}"$

Then the characteristic polynomial is as follows.

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

Solve the characteristic polynomial and find the roots as follows.

$\begin{array}{l}{\mathrm{\lambda }}^2=0\\ \mathrm{\lambda }=0\\ \mathrm{\lambda }=0,\mathrm{\lambda }=0\end{array}$

Therefore, the roots of the characteristic polynomials are 0 and 0.

Conclusion of the solution

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

The exponential functions${\mathrm{e}}^{0\mathrm{t}}=1$and${\mathrm{te}}^{0\mathrm{t}}=\mathrm{t}$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.

Thus, the general solution of the differential equation $f\text{'}\left(t\right)=0\mathrm{is}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1+{\mathrm{C}}_2\mathrm{t}$.