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 by $\mathbf{渭}\left(x\right)$the formula .
• Multiply the equation in standard form by $\mathbf{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$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

$$\begin{gathered} \mathit{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathit{P}\mathit{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{y}\mathbf{=}\mathit{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathit{渭}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

Given that, ${\mathrm{RA}}_1$decays into ${\mathrm{RA}}_2$is and the decay constant for $40e^{-20t}\mathrm{kg}/\sec$is $\mathrm{k}=5/\sec$

That is, $\frac{\mathrm{dA}}{\mathrm{dt}}=40{\mathrm{e}}^{-20\mathrm{t}}\mathrm{kg}/\sec ;\mathrm{k}=5/\sec$.

To find the mass of ${\mathrm{RA}}_2$for$t鈮�0$if initially$y\left(0\right)=10\mathrm{kg}$

Let us create an initial value problem,

$\frac{\mathrm{dy}}{\mathrm{dt}}+5\mathrm{y}=40{\mathrm{e}}^{-20\mathrm{t}},\mathrm{y}\left(0\right)=10\mathrm{kg}......\left(1\right)$

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

Where role="math" localid="1664172714912" $\mathrm{P}\left(t\right)=5$.

Then,

$\begin{array}{rcl}\mathrm{渭}\left(\mathrm{t}\right)& =& {\mathrm{e}}^{鈭�\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{鈭�5\mathrm{tdt}}\\ & =& {\mathrm{e}}^{5\mathrm{t}}\end{array}$

Multiply in equation (1)

$$\begin{gathered} e^{5t}\frac{\mathrm{dy}}{\mathrm{dt}}+5e^{5t}\mathrm{y}=40e^{5t}{\mathrm{e}}^{-20\mathrm{t}} \\ {\mathrm{e}}^{5\mathrm{t}}\frac{\mathrm{dy}}{\mathrm{dt}}+5{\mathrm{e}}^{5\mathrm{t}}\mathrm{y}=40{\mathrm{e}}^{-15\mathrm{t}} \\ \frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{e}}^{5\mathrm{t}}\mathrm{y}\right]=40{\mathrm{e}}^{-15\mathrm{t}} \end{gathered}$$

Integrating both sides.

$$\begin{gathered} {鈭�}_{}^{}\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{e}}^{5\mathrm{t}}\mathrm{y}\right]dt={鈭�}_{}^{}40{\mathrm{e}}^{-15\mathrm{t}}\mathrm{dt} \\ {\mathrm{e}}^{5\mathrm{t}}\mathrm{y}=-\frac{40}{15}{\mathrm{e}}^{-15\mathrm{t}}+\mathrm{C} \\ \mathrm{y}=-\frac{8}{3}\frac{{\mathrm{e}}^{-15\mathrm{t}}}{{\mathrm{e}}^{5\mathrm{t}}}+\frac{\mathrm{C}}{{\mathrm{e}}^{5\mathrm{t}}} \\ \mathrm{y}=-\frac{8}{3}{\mathrm{e}}^{-20\mathrm{t}}+{\mathrm{Ce}}^{-5\mathrm{t}} \end{gathered}$$

So, the solution is $\mathrm{y}=-\frac{8}{3}{\mathrm{e}}^{-20\mathrm{t}}+{\mathrm{Ce}}^{-5\mathrm{t}}.......\left(2\right)$

Given that, $y\left(0\right)=10\mathrm{kg}$.

Then, t = 0 and y = 10.

Substitute the value in equation (2) to get the value of C.

role="math" localid="1664173608266" $\begin{array}{rcl}\mathrm{y}& =& -\frac{8}{3}{\mathrm{e}}^{-20\mathrm{t}}+{\mathrm{Ce}}^{-5\mathrm{t}}\\ 10& =& -\frac{8}{3}{\mathrm{e}}^{-20\left(0\right)}+{\mathrm{Ce}}^{-5\left(0\right)}\\ 10& =& -\frac{8}{3}+\mathrm{C}\\ \mathrm{C}& =& \frac{38}{3}\end{array}$

Substitute the value of C in equation (2).

$\begin{array}{rcl}\mathrm{y}& =& -\frac{8}{3}{\mathrm{e}}^{-20\mathrm{t}}+\frac{38}{3}{e}^{-5\mathrm{t}}\end{array}$

So, the solution is$\begin{array}{rcl}\mathbf{y}& \mathbf{=}& \mathbf{-}\frac{\mathbf{8}}{\mathbf{3}}{\mathbf{e}}^{\mathbf{-}\mathbf{20}\mathbf{t}}\mathbf{+}\frac{\mathbf{38}}{\mathbf{3}}{\mathit{e}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{}\end{array}$