Question

Verifying solutions of initial value problems Verify that the given function \(y\) is a solution of the initial value problem that follows it. $$y(x)=\frac{1}{4}\left(e^{2 x}-e^{-2 x}\right) ; y^{\prime \prime}(x)-4 y(x)=0, y(0)=0, y^{\prime}(0)=1$$

Short answer

Answer: Yes, the function \(y(x) = \frac{1}{4}(e^{2x} - e^{-2x})\) is a solution to the given initial value problem, since it satisfies both the differential equation and the initial conditions.

Step-by-step solution

Compute the first and second derivatives of the function

To check if the function satisfies the given ODE, we need to find its first and second derivatives. Given \(y(x)=\frac{1}{4}(e^{2x}-e^{-2x})\), we can calculate the derivatives as follows: First derivative: $$y'(x) = \frac{1}{4}(2e^{2x}+2e^{-2x})$$ Second derivative: $$y''(x) = \frac{1}{4}(4e^{2x}-4e^{-2x})$$

Plug the derivatives into the ODE

Now we need to plug the second derivative \(y''(x)\) and the function \(y(x)\) into the ODE to check if it holds true: $$y''(x) - 4y(x) = 0$$ Substitute the expressions we found in Step 1 into this equation: $$\frac{1}{4}(4e^{2x}-4e^{-2x}) - 4\cdot\frac{1}{4}(e^{2x}-e^{-2x}) = 0$$ Simplify the equation: $$(e^{2x}-e^{-2x}) - (e^{2x}-e^{-2x}) = 0$$ The equation holds, as both sides are equal.

Verify the initial conditions

We need to check if the function and its derivative satisfy the initial conditions: $$y(0) = 0, y'(0) = 1$$ Plug \(x=0\) into \(y(x)\): $$y(0) = \frac{1}{4}(e^{0}-e^{0}) = 0$$ Plug \(x=0\) into \(y'(x)\): $$y'(0) = \frac{1}{4}(2e^{0}+2e^{0}) = 1$$ Both initial conditions are satisfied.

Conclusion

Since the given function, its derivatives and initial conditions satisfy the requirements of the initial value problem, we can conclude that \(y(x)=\frac{1}{4}(e^{2x}-e^{-2x})\) is a solution to the given initial value problem: $$y''(x)-4y(x)=0, y(0)=0, y'(0)=1$$

Key concepts

Differential Equations

Differential equations are powerful tools in mathematics and physics, representing relationships between functions and their derivatives. In essence, they describe how a particular quantity changes over time or space. An ordinary differential equation (ODE) focuses on functions of a single variable and their derivatives. The exercise presents an ODE, namely, y''(x) - 4y(x) = 0, which is a second-order linear homogeneous differential equation. It's 'homogeneous' because the equation is set to zero, and 'linear' because both y and its derivatives appear to the first power without any products between them.

Understanding differential equations is crucial for various applications, including physics, engineering, and economics, as they can model everything from population growth to the motion of celestial bodies. The main challenge lies in finding a function, or a 'solution,' that satisfies the given equation, which can necessitate techniques ranging from separation of variables to Fourier series in more complex cases.

Derivative Calculations

Derivatives represent the rate at which a function is changing at any point, and they are foundational to calculus and differential equations. Solving an ODE typically requires calculating the first, second, or even higher-order derivatives of the unknown function. In the textbook exercise, the first derivative y'(x) is obtained by applying the chain rule to the exponential functions, while the second derivative y''(x) comes from differentiating the first derivative.

Mastering derivative calculations involves learning various rules such as the product rule, quotient rule, and chain rule. Beyond solving ODEs, derivatives are used in optimizing functions to find maximum and minimum values, which reflects their significance in fields like economics, biology, and engineering.

Solving ODEs

Solving ordinary differential equations (ODEs) is a multi-step process of finding a function that satisfies the relationship outlined in the differential equation. Initial value problems add specific conditions that the solution must meet at certain points, which help determine the constants that may arise during the solving process. The exercise given is a classic example of an initial value problem, where the solution y(x) must satisfy not only the ODE but also the conditions y(0) = 0 and y'(0) = 1.

Techniques to solve ODEs vary. Some, like separation of variables, work well for first-order ODEs. Others, like the method of undetermined coefficients or variation of parameters, are ideal for linear ODEs with constant coefficients. The goal is always the same: find a function that satisfies both the differential equation and the initial conditions.

Exponential Functions

Exponential functions, such as e^{2x} and e^{-2x}, appear frequently in differential equations and have distinctive characteristics. An exponential function is of the form y = e^{kx}, where e (approximately 2.71828) is the base of natural logarithms, k is a constant, and x is the exponent. These functions grow (or decay) at rates proportional to their values, which is why they naturally describe processes involving growth or decay, like radioactive decay or population growth.

In the context of the textbook exercise, understanding the behavior of exponential functions helps to verify that the proposed solution meets the ODE. Characteristics of exponential functions, such as their derivatives, are themselves exponential functions, play a pivotal role in examining how the initial value problem's solution evolves and confirms its validity.