Question

Radioactive Decay Series The following system of differential equations is encountered in the study of the decay of a special type of radioactive series of elements :$\begin{array}{l}\frac{dx}{dt}=-{\lambda }_{1}x\\ \frac{dy}{dt}={\lambda }_{1}x-{\lambda }_{2}y\end{array}$

Where ${\lambda }_{1}and{\lambda }_2$ are constants .Discuss how to solve this system subject to $x\left(0\right)=x_0,y\left(0\right)=y_0$.Carry out your ideas

Short answer

The value of x and y

$$\begin{gathered} x=x_0{e}^{-l_{1}t} \\ y=\frac{x_0{l}_1}{l_2-l_1}{e}^{-l_{1}t}+\left(y_0-\frac{x_0{l}_1}{l_2-l_1}\right)e^{-I_{2}t} \end{gathered}$$

Step-by-step solution

Definition of initial-value problem

The interval I, containing $x_0$ the problem solving an nth order differential equations subjected to n side conditions specified at x₀

Solve:$\frac{{\mathrm{d}}^{\mathrm{n}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{n}}}=\mathrm{F}(\mathrm{x},\mathrm{y},\mathrm{y}\text{'},.....,{\mathrm{y}}^{(\mathrm{n}-1)})$

Subject to: $y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,\mathrm{...},y^{(n-1)}\left(x_0\right)=y_{n-1}$

Where, $y_0,y_1,\mathrm{.....},y_{n-1}$are arbituary constants called an nth-order initial value problem (IVP)

The values of y(x) and its first (n-1) derivatives at $y_0,y\text{'}\left(x_0\right)=y_1,\mathrm{...},y^{(n-1)}\left(x_0\right)=y_{n-1}$are called initial conditions(IC)

Determination of first differential equation solution 

The given constants are ${\lambda }_1$and ${\lambda }_2$;$x\left(0\right)=x_0,y\left(0\right)=y_0$

$\begin{array}{l}\frac{\mathrm{dx}}{\mathrm{dt}}=-{\mathrm{\lambda }}_1\mathrm{x}\\ \frac{\mathrm{dy}}{\mathrm{dt}}={\mathrm{\lambda }}_1\mathrm{x}-{\mathrm{\lambda }}_2\mathrm{y}\end{array}$

Solve the first differential equation ,and then substitute the first solution in second differential equation.

$\begin{array}{l}\frac{1}{x}dx=-{\lambda }_{1}dt\\ \int \frac{1}{x}dx=\int -{\lambda }_{1}dt\\ \ln x=-{\lambda }_{1}t+C\end{array}$

Solving first differential equation for value C

$\begin{array}{l}\ln x_0=0+C\\ C=\ln x_0\end{array}$

Sub initial condition$x\left(0\right)=x_0$

Then;

$\begin{array}{l}\ln x=-{\lambda }_{1}t+\ln x_0\\ \ln \frac{x}{x_0}=-{\lambda }_{1}t\\ \frac{x}{x_0}=e^{-{\lambda }_{1}t}\\ x=x_0{e}^{-{\lambda }_{1}t}\end{array}$

It is a final solution for first differential eqn.

Determination of second differential equation solution

$\frac{dy}{dt}={\lambda }_1\left(x_0{e}^{-{\lambda }_{1}t}\right)-{\lambda }_{2}y$

Sub solution in second differential eqn

$$\begin{gathered} \frac{dy}{dt}+l_{2}y=l_1\left(x_0{e}^{-l_{1}t}\right) \\ \frac{dy}{dt}+l_{2}y=l_1\left(x_0{e}^{-l_{1}t}\right) \\ e{\int }^{I_{2}dt}=e^{I_{2}t} \end{gathered}$$

Now solving to get the integrating factor

……..multiplying both sides by integrating factor

$$\begin{gathered} e^{I2t}\left[\frac{dy}{dt}+l_{2}y\right]=e^{I2t}\left[x_0{l}_1{e}^{-l_{1}t}\right] \\ \frac{d}{dt}=x_0{l}_1{e}^{\left(I_2-I_1\right)t} \\ \int d\left[e^{I2t}y\right]=\int x_0{l}_1{e}^{\left(I_2-I_1\right)t} \end{gathered}$$

Now

$\begin{array}{l}{y}_0=\frac{x_0{\lambda }_1}{{\lambda }_2-{\lambda }_1}+C\\ C=y_0-\frac{x_0{\lambda }_1}{{\lambda }_2-{\lambda }_1}\end{array}$

Put$y\left(0\right)=y_0$

$y=\frac{x_0{l}_1}{l_2-l_1}{e}^{{\lambda }_{1}t}+\left(y_0-\frac{x_0{l}_1}{l_2-l_1}\right)e^{I_{2}t}$

Thus,it is the final solution of second differential equation