91影视

• Write the equation in the standard form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.
• Calculate the integrating factor $\mathit{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$by the formula $\mathit{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{e}\mathit{x}\mathit{p}\mathbf{\left[}{\mathbf{鈭�}}_{}^{}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dx}\mathbf{\right]}$.
• Multiply the equation in standard form by $\mathit{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and, recalling that the left-hand side is just $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}$, obtain

Given that,

$\frac{\mathrm{dy}}{\mathrm{dx}}+4\mathrm{y}-{\mathrm{e}}^{\mathrm{x}}=0,\mathrm{y}\left(0\right)=\frac{4}{3}......\left(1\right)$

Write the equation (1) into standard form of linear equation.

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}+4y-e^{-\mathrm{x}}=0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}+4y=e^{-\mathrm{x}}........\left(2\right) \end{gathered}$$

Calculate the integrating factor of $渭\left(x\right)$.

Where $\mathrm{P}\left(x\right)=4$.

Then,

$\begin{array}{rcl}\mathrm{渭}\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{鈭�}_{}^{}\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{{鈭�}_{}^{}4\mathrm{xdx}}\\ & =& {\mathrm{e}}^{4\mathrm{x}}\end{array}$

Multiply $渭\left(x\right)$in equation (2)

$\begin{array}{rcl}{\mathrm{e}}^{4\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}+4{\mathrm{e}}^{4\mathrm{x}}\mathrm{y}& =& {\mathrm{e}}^{4\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\\ {\mathrm{e}}^{4\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}+4{\mathrm{e}}^{4\mathrm{x}}\mathrm{y}& =& {\mathrm{e}}^{3\mathrm{x}}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{e}}^{4\mathrm{x}}\mathrm{y}\right]& =& {\mathrm{e}}^{3\mathrm{x}}\end{array}$

Integrating both sides.

role="math" localid="1664123416648" $\begin{array}{rcl}{鈭�}_{}^{}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{e}}^{4\mathrm{x}}\mathrm{y}\right]dx& =& 鈭�{\mathrm{e}}^{3\mathrm{x}}\mathrm{dx}\\ {\mathrm{e}}^{4\mathrm{x}}\mathrm{y}& =& \frac{1}{3}{\mathrm{e}}^{3\mathrm{x}}+\mathrm{C}\\ \mathrm{y}& =& \frac{1}{3{\mathrm{e}}^{\mathrm{x}}}+{\mathrm{Ce}}^{-4\mathrm{x}}........\left(3\right)\end{array}$

$y\left(0\right)=\frac{4}{3}$Given that, $y\left(0\right)=\frac{4}{3}$.

Then, $x=0$and $y\left(0\right)=\frac{4}{3}$.

Substitute the value in equation (3) to get the value of C.

$\begin{array}{rcl}\mathrm{y}& =& \frac{1}{3{\mathrm{e}}^{\mathrm{x}}}+{\mathrm{Ce}}^{-4\mathrm{x}}\\ \frac{4}{3}& =& \frac{1}{3{\mathrm{e}}^0}+{\mathrm{Ce}}^0\\ C& =& \frac{4}{3}-\frac{1}{3}\\ C& =& 1\end{array}$

Substitute the value of C in equation (2).

$\begin{array}{rcl}\mathrm{y}& =& \frac{1}{3{\mathrm{e}}^{\mathrm{x}}}+e^{-4\mathrm{x}}\end{array}$

So, the solution is $\begin{array}{rcl}\mathbf{y}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{3}{\mathbf{e}}^{\mathbf{x}}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{4}\mathbf{x}}\mathbf{}\end{array}$