Question

Solve the initial value problem in$\mathbf{f}\mathbf{\text{'}}\left(t\right)\mathbf{+}\mathbf{2}\mathbf{f}\left(t\right)\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{f}\left(1\right)\mathbf{=}\mathbf{1}$

Short answer

The solution is.$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}+2}$

Step-by-step solution

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 g(t) is a smooth function and 'a' is a constant. Then the general solution will be role="math" localid="1660806053060" $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{at}}\mathbf{\int }{\mathbf{e}}^{\mathbf{-}\mathbf{at}}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$.

Determination of the solution

Consider the differential equation as follows.

$\mathrm{f}\text{'}\left(\mathrm{t}\right)‐2\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}$

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${\mathrm{e}}^{2\mathrm{t}}$ for g(t) and 2 for f (t) 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.

role="math" localid="1660806345559" $$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\int {\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2t}\int {\mathrm{e}}^{2t}\times 0\times \mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\int 0\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\mathrm{C} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\mathrm{C} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\left(\mathrm{C}\right) \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\mathrm{C}+{\mathrm{e}}^2\mathrm{C} \end{gathered}$$

Hence, the solution for the linear differential equation $\mathrm{f}\text{'}\left(\mathrm{t}\right)+2\mathrm{f}\left(\mathrm{t}\right)=0$is.

role="math" localid="1660806669972" $$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\mathrm{C}+{\mathrm{e}}^2\mathrm{C} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}+2}\mathrm{C} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}+2}\left(\because \mathrm{C}=1\right) \end{gathered}$$

Thus, the solution is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}+2}$.