91影视

If $位$is a complex number, then$z=e^{位t}$ is the unique complex valued function such that

$\frac{dz}{dt}=位t$and$z\left(0\right)=1$

Consider the differential equation$\mathbf{f}\mathbf{\text{'}}\left(t\right)\mathbf{鈥�}\mathbf{af}\left(t\right)\mathbf{=}\mathbf{g}\left(t\right)$where$\mathbf{g}\left(t\right)$is a smooth function and$\mathbf{\text{'}}\mathbf{a}\mathbf{\text{'}}$is a constant. Then the general solution will be$\mathbf{f}\left(t\right)\mathbf{=}{\mathbf{e}}^{\mathbf{at}}\mathbf{鈭�}{\mathbf{e}}^{\mathbf{-}\mathbf{at}}\mathbf{g}\left(t\right)\mathbf{dt}$.

Consider the differential equation as follows.

$\mathrm{f}\text{'}\left(\mathrm{t}\right)鈥�5\mathrm{f}\left(\mathrm{t}\right)=0$

Now, the differential equation is in the form as follows.

$\mathrm{f}\text{'}\left(\mathrm{t}\right)鈥�\mathrm{af}\left(\mathrm{t}\right)=\mathrm{g}\left(\mathrm{t}\right)$, where$\mathrm{g}\left(\mathrm{t}\right)$is a smooth function, then the general solution will be as follows.

$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}鈭�{\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$

Substitute the value 0 for g(t) and 5 for a in $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}鈭�{\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}鈭�{\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{5\mathrm{t}}鈭�{\mathrm{e}}^{-5\mathrm{t}}脳0脳\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{5\mathrm{t}}鈭�0脳\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{5\mathrm{t}}路\mathrm{C} \end{gathered}$$

Hence, the solution for the linear differential equation $\mathrm{f}\text{'}\left(\mathrm{t}\right)鈥�5\mathrm{t}\left(\mathrm{t}\right)=0$is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{Ce}}^{5\mathrm{t}}$.

Consider the initial value problem as$f\text{'}\left(t\right)-5f\left(t\right)=0;f\left(0\right)=3$

Since the initial value problem $f\text{'}\left(t\right)-5f\left(t\right)=0;f\left(0\right)=3$is first order homogeneous equation.

Then the solution becomes

$$\begin{gathered} f\text{'}\left(t\right)-5f\left(t\right)=0;f\left(0\right)=3 \\ T\left(f\right)=0 \\ f\text{'}\left(t\right)-5f\left(t\right)=0 \\ f\text{'}\left(t\right)=5f\left(t\right) \end{gathered}$$

By definition of an exponential functions, the solution are the functions of the form

$f\left(t\right)=C{e}^{at}$,here C is an arbitrary function

Here C=3 and a=5 for the solution of the function

Thus $f\left(t\right)=3e^{5t}$.

Hence, the solution.