Problem 14 - Advanced Engineering Mathematics

Question

y=c_{1} e^{x}+c_{2} e^{-x}\( is a two-parameter family of solutions of the second-order DE \)y^{\prime \prime}-y=0$. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. $$ y(0)=0, \quad y^{\prime}(0)=0 $$

Step-by-Step Solution

Answer

The specific solution is \( y(x) = 0 \).

Step 1: Understand the General Solution

The given differential equation is \( y'' - y = 0 \). The general solution is provided as \( y = c_1 e^x + c_2 e^{-x} \) where \( c_1 \) and \( c_2 \) are constants to be determined based on initial conditions.

Step 2: Apply Initial Condition - Position

The initial condition \( y(0) = 0 \) means if we substitute \( x = 0 \) in the equation, we get: \[ c_1 e^0 + c_2 e^0 = c_1 + c_2 = 0 \]This equation can be rearranged to find the relationship between \( c_1 \) and \( c_2 \).

Step 3: Derive and Apply Initial Condition - Velocity

The derivative of the general solution is \( y' = c_1 e^x - c_2 e^{-x} \). Applying the initial condition \( y'(0) = 0 \) gives: \[ c_1 e^0 - c_2 e^0 = c_1 - c_2 = 0 \]

Step 4: Solve Simultaneous Equations

Now, solve the simultaneous equations: 1. \( c_1 + c_2 = 0 \)2. \( c_1 - c_2 = 0 \)Adding these equations, we get \( 2c_1 = 0 \), thus \( c_1 = 0 \). Similarly, \( c_2 = 0 \).

Step 5: Write the Specific Solution

Substituting \( c_1 = 0 \) and \( c_2 = 0 \) into the general solution gives \( y = 0 \). Therefore, the specific solution satisfying the given initial conditions is \( y(x) = 0 \).

Step 6: Conclusion

The solution to the initial value problem is the trivial solution \( y(x) = 0 \). This satisfies both the differential equation and the initial conditions provided.

Key Concepts

Initial Value Problems

Initial value problems (IVPs) are a fundamental concept in the study of differential equations. They involve finding a function that satisfies a given differential equation, along with specified initial conditions. These initial conditions allow us to determine unique solutions by providing specific values for the function or its derivatives at a particular point.In the context of our problem, we were given the differential equation \( y'' - y = 0 \) and two initial conditions: \( y(0) = 0 \) and \( y'(0) = 0 \). These conditions are applied at \( x = 0 \), which is our starting point or the 'initial' point. By using these conditions, we determine the constants \( c_1 \) and \( c_2 \) in the general solution \( y = c_1 e^x + c_2 e^{-x} \).The beauty of IVPs is that they allow us to hone in on one particular solution from a potentially infinite family of solutions. Without initial conditions, the general solution remains too broad and unspecific. Thus, initial conditions are crucial for solving real-world problems where specific outcomes are desired.

Second-order Differential Equations

Second-order differential equations are equations involving the second derivative of a function. These are more complex than first-order differential equations as they take into account the rate of change of the rate of change of the function.In our exercise, the differential equation given is \( y'' - y = 0 \). This is a homogeneous second-order linear differential equation, as it depends only on \( y'' \) and \( y \) without any additional terms or functions. Such equations often take the form of \( ay'' + by' + cy = 0 \), but here, \( a = 1 \), \( b = 0 \), and \( c = -1 \). The solutions to these equations commonly involve exponential functions because the exponential function retains its form upon differentiation. The general solution provided \( y = c_1 e^x + c_2 e^{-x} \) reflects the nature of solutions to these types of differential equations, with the constants \( c_1 \) and \( c_2 \) revealed through initial conditions.

Exponential Functions

Exponential functions are critical in solving differential equations due to their unique properties. An exponential function is of the form \( e^x \) or \( e^{-x} \), where \( e \) is the base of the natural logarithm. These functions are notable because their rate of growth or decay is proportional to their value, a property that makes them ideal solutions for differential equations.In our specific problem, exponential functions \( e^x \) and \( e^{-x} \) are used in the general solution \( y = c_1 e^x + c_2 e^{-x} \) of the differential equation \( y'' - y = 0 \). The advantage of exponential functions in this context is that they remain exponential after differentiation, a key attribute when dealing with second-order differential equations.Upon differentiation, the form \( y' = c_1 e^x - c_2 e^{-x} \) arises naturally, showing how the exponential terms play a direct role in forming the relationship between the function and its derivatives. Consequently, exponential functions are indispensable tools in the toolkit for solving differential equations.