Problem 1 - Advanced Engineering Mathematics

Question

The linear equation \(d x / d t=-\lambda_{1} x\) can be solved by either separation of variables or by an integrating factor. Integrating both sides of \(d x / x=-\lambda_{1} d t\) we obtain \(\ln |x|=-\lambda_{1} t+c\) from which we get \(x=c_{1} e^{-\lambda_{1} t} .\) Using \(x(0)=x_{0}\) we find \(c_{1}=x_{0}\) so that \(x=x_{0} e^{-\lambda_{1} t} .\) Substituting this result into the second differential equation we have $$\frac{d y}{d t}+\lambda_{2} y=\lambda_{1} x_{0} e^{-\lambda_{1} t}$$ which is linear. An integrating factor is \(e^{\lambda_{2} t}\) so that $$\frac{d}{d t}\left[e^{\lambda_{2} t} y\right]=\lambda_{1} x_{0} e^{\left(\lambda_{2}-\lambda_{1}\right) t}+c_{2}$$ $$y=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{\left(\lambda_{2}-\lambda_{1}\right) t} e^{-\lambda_{2} t}+c_{2} e^{-\lambda_{2} t}=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+c_{2} e^{-\lambda_{2} t}$$ Using \(y(0)=0\) we find \(c_{2}=-\lambda_{1} x_{0} /\left(\lambda_{2}-\lambda_{1}\right) .\) Thus $$y=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}}\left(e^{-\lambda_{1} t}-e^{-\lambda_{2} t}\right)$$ Substituting this result into the third differential equation we have $$\frac{d z}{d t}=\frac{\lambda_{1} \lambda_{2} x_{0}}{\lambda_{2}-\lambda_{1}}\left(e^{-\lambda_{1} t}-e^{-\lambda_{2} t}\right)$$ Integrating we find $$z=-\frac{\lambda_{2} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{2} t}+c_{3}$$ Using \(z(0)=0\) we find \(c_{3}=x_{0} .\) Thus $$z=x_{0}\left(1-\frac{\lambda_{2}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{2} t}\right)$$.

Step-by-Step Solution

Answer

Solve the ODEs using separation of variables and integrating factors, resulting in functions for \(x\), \(y\), and \(z\).

Step 1: Solve the First Differential Equation

The first differential equation is \( \frac{dx}{dt} = -\lambda_1 x \). We can solve it by separation of variables: rearrange to get \( \frac{dx}{x} = -\lambda_1 dt \). Integrate both sides to find \( \ln|x| = -\lambda_1 t + c \). Exponentiating gives \( x = c_1 e^{-\lambda_1 t} \). Using the initial condition \( x(0) = x_0 \), we find \( c_1 = x_0 \), hence \( x = x_0 e^{-\lambda_1 t} \).

Step 2: Formulate the Second Differential Equation

Substitute \( x = x_0 e^{-\lambda_1 t} \) into the second equation \( \frac{dy}{dt} + \lambda_2 y = \lambda_1 x_0 e^{-\lambda_1 t} \). This is a linear first-order differential equation.

Step 3: Solve the Second Differential Equation Using an Integrating Factor

The integrating factor is \( e^{\lambda_2 t} \). Multiplying through by this gives \( \frac{d}{dt} [e^{\lambda_2 t} y] = \lambda_1 x_0 e^{(\lambda_2 - \lambda_1)t} \). Integrating both sides yields \( y = \frac{\lambda_1 x_0}{\lambda_2 - \lambda_1} e^{-\lambda_1 t} + c_2 e^{-\lambda_2 t} \). Using the condition \( y(0) = 0 \), we solve for \( c_2 \) to find \( c_2 = -\frac{\lambda_1 x_0}{\lambda_2 - \lambda_1} \), resulting in \( y = \frac{\lambda_1 x_0}{\lambda_2 - \lambda_1} (e^{-\lambda_1 t} - e^{-\lambda_2 t}) \).

Step 4: Solve the Third Differential Equation

Substitute \( y = \frac{\lambda_1 x_0}{\lambda_2 - \lambda_1} (e^{-\lambda_1 t} - e^{-\lambda_2 t}) \) into the third equation \( \frac{dz}{dt} = \frac{\lambda_1 \lambda_2 x_0}{\lambda_2 - \lambda_1} (e^{-\lambda_1 t} - e^{-\lambda_2 t}) \). Integrate to find \( z = -\frac{\lambda_2 x_0}{\lambda_2 - \lambda_1} e^{-\lambda_1 t} + \frac{\lambda_1 x_0}{\lambda_2 - \lambda_1} e^{-\lambda_2 t} + c_3 \). Using \( z(0) = 0 \), solve for \( c_3 \) and find \( c_3 = x_0 \), so finally \( z = x_0(1 - \frac{\lambda_2}{\lambda_2 - \lambda_1} e^{-\lambda_1 t} + \frac{\lambda_1}{\lambda_2 - \lambda_1} e^{-\lambda_2 t}) \).

Key Concepts

Separation of Variables

When solving differential equations, one of the simplest and most frequently used methods is separation of variables. The idea is to rearrange the equation so that each variable appears on a different side of the equation. This way, we are able to integrate each side separately.

For example, take the differential equation \( \frac{dx}{dt} = -\lambda_1 x \). Here, we want to separate \( x \) and \( t \). This can be done by rearranging it to \( \frac{dx}{x} = -\lambda_1 dt \). Now, each variable is on its own side.

• Integrate both sides: The left side becomes \( \ln|x| \) and the right side just becomes a simple integral \( -\lambda_1 t + c \).
  
• Use algebra to solve for \( x \), resulting in \( x = c_1 e^{-\lambda_1 t} \).
  

Through this method, separation of variables allows you to solve many differential equations with ease.

Integrating Factor

Another robust technique for solving differential equations is called the Integrating Factor. This method is particularly useful when dealing with linear first-order differential equations, those having the standard form \( \frac{dy}{dt} + P(t)y = Q(t) \).

The integrating factor, which is a function, helps simplify the equation in such a way that it becomes easy to integrate directly. Let's see how it works in action for the second differential equation:

The equation \( \frac{dy}{dt} + \lambda_2 y = \lambda_1 x_0 e^{-\lambda_1 t} \) is linear. To solve it, first identify the integrating factor, which in this case is \( e^{\lambda_2 t} \). This multiplying factor aims to transform the left-hand side into the derivative of a product. Here's what to do:

• Multiply every term in the equation by the integrating factor: \( e^{\lambda_2 t} \cdot \frac{dy}{dt} + e^{\lambda_2 t} \lambda_2 y = \lambda_1 x_0 e^{(\lambda_2 - \lambda_1)t} \).
  
• Notice that this simplifies to \( \frac{d}{dt} [e^{\lambda_2 t} y] \).
  
• Integrate both sides with respect to \( t \), simplifying the integration procedure.
  

This technique effectively converts a difficult problem into a more straightforward one, allowing integration straight away.

Initial Value Problem

An initial value problem (IVP) provides an extra layer of specificity to the solution of a differential equation. Beyond just finding a general solution, the aim here is to find the particular solution that meets given conditions (initial values).

To understand an IVP better, consider that in any differential equation, you'll typically end up with one or more constants, like \( c_1, c_2, \) etc., after integration. An IVP provides additional information to find these specific constants.

For instance, when solving \( \frac{dx}{dt} = -\lambda_1 x \), if we know \( x(0)=x_0 \), then we can replace \( t=0 \) in the solution \( x = c_1 e^{ -\lambda_1 t } \) to find \( c_1 \) specifically. This transforms it to \( x_0 = c_1 \cdot 1 \), giving \( c_1 = x_0 \).

• IVPs provide real-world linkage by fixing the arbitrary constants in your solution.
  
• They allow you to predict the behavior of a system under specified starting conditions.
  

This is crucial because many applications need precise solutions rather than a general form.