Question

Solve the initial value problem in$\mathbf{f}\mathbf{\text{'}}\left(t\right)\mathbf{‐}\mathbf{5}\mathbf{f}\left(t\right)\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{f}\left(0\right)\mathbf{=}\mathbf{3}$

Short answer

The solution is.$\mathrm{f}\left(\mathrm{t}\right)=3{\mathrm{e}}^{5\mathrm{t}}$

Step-by-step solution

Definition for the Complex Exponential functions

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

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

Definition of first order linear differential equation

Consider the differential equation$\mathbf{f}\mathbf{\text{'}}\left(t\right)\mathbf{‐}\mathbf{af}\left(t\right)\mathbf{=}\mathbf{g}\left(t\right)$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$\mathbf{f}\left(t\right)\mathbf{=}{\mathbf{e}}^{\mathbf{at}}\mathbf{\int }{\mathbf{e}}^{\mathbf{-}\mathbf{at}}\mathbf{g}\left(t\right)\mathbf{dt}$.

Determination of the solution

Consider the differential equation as follows.

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

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

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

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

Compute the calculation of the solution

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

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\int {\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{5\mathrm{t}}\int {\mathrm{e}}^{-5\mathrm{t}}\times 0\times \mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{5\mathrm{t}}\int 0\times \mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{5\mathrm{t}}·\mathrm{C} \end{gathered}$$

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

Explanation for the solution of the initial value problem f'(t)-5f(t)=0;f(0)=3

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.