Question

Solve the initial value problem in $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{=}\mathbf{7}\mathbf{,}\mathbf{x}\left(0\right)\mathbf{=}\mathbf{0}$

Short answer

The solution is. $\mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3}$.

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$\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ is a smooth function and 'a' 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.

$\mathrm{x}\text{'}\left(\mathrm{t}\right)+3\mathrm{x}\left(\mathrm{t}\right)=7$

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.

role="math" localid="1660803566306" $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\int {\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$

Compute the calculation of the solution.

Substitute the value 7 for g(t) and $‐3$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.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\int {\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-3t}\int {\mathrm{e}}^{3t}\times 7\times \mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)=7{\mathrm{e}}^{-3\mathrm{t}}\int {\mathrm{e}}^{3\mathrm{t}}\mathrm{dt} \end{gathered}$$

Using substitution method in the integral as follows.

$$\begin{gathered} \mathrm{y}=3\mathrm{t} \\ \mathrm{dy}=3\mathrm{dt} \\ \frac{\mathrm{dy}}{3}=\mathrm{dt} \end{gathered}$$

Substitute the value 3t for y and $\frac{\mathrm{dy}}{3}$for dt in role="math" localid="1660804416396" $\mathrm{f}\left(\mathrm{t}\right)=7{\mathrm{e}}^{3t}\int {\mathrm{e}}^{-3t}\mathrm{dt}$as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)=7{\mathrm{e}}^{3\mathrm{t}}\int {\mathrm{e}}^{-3\mathrm{t}}\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)=7{\mathrm{e}}^{3\mathrm{t}}\int {\mathrm{e}}^{\mathrm{y}}\frac{\mathrm{dy}}{3} \\ \mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3}{\mathrm{e}}^{3\mathrm{t}}\left({\mathrm{e}}^{\mathrm{y}}+\mathrm{C}\right) \end{gathered}$$

Now, undo the substitution as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3}{\mathrm{e}}^{-3\mathrm{t}}\left({\mathrm{e}}^{\mathrm{y}}+\mathrm{C}\right) \\ \mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3}{\mathrm{e}}^{-3\mathrm{t}}\left({\mathrm{e}}^{3\mathrm{t}}+\mathrm{C}\right) \\ =\frac{7}{3}{\mathrm{e}}^{-3\mathrm{t}}{\mathrm{e}}^{3\mathrm{t}}+\frac{7}{3}{\mathrm{e}}^{-3\mathrm{t}}\mathrm{C} \\ \mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3}+\frac{7}{3}{\mathrm{e}}^{-3\mathrm{t}}\mathrm{C} \end{gathered}$$

Hence, the solution for the linear differential equation $\mathrm{x}\text{'}\left(\mathrm{t}\right)+3\mathrm{x}\left(\mathrm{t}\right)=7$is $\mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3}+\frac{7}{3}{\mathrm{e}}^{-3\mathrm{t}}\mathrm{C}$.

Explanation for the solution of the initial value problem dxdt+3x=7,x(0)=0

Consider the initial value problem as $\frac{dx}{dt}+3x=7,x\left(0\right)=0$

Since the initial value problem $\frac{dx}{dt}+3x=7,x\left(0\right)=0$is first order homogeneous equation.

Then the solution becomes

role="math" localid="1660804903525" $\mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3}+\frac{7}{3}{\mathrm{e}}^{-3\mathrm{t}}\mathrm{C}$

Now substitute the initial conditions we get

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3}+\frac{7}{3}{\mathrm{e}}^{-3\mathrm{t}}\mathrm{C} \\ \mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3}+\frac{7}{3}{\mathrm{e}}^{-3\mathrm{t}}\left(0\right) \\ \mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3}+0 \\ \mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3} \end{gathered}$$

Thus the solution is $\mathrm{f}\left(\mathrm{t}\right)=\frac{7}{3}$.