Question

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

Short Answer: Yes, the function \(y(x)=\frac{1}{4}\left(e^{2 x}-e^{-2 x}\right)\) is a solution to the initial value problem with the differential equation \(y''(x)-4y(x)=0\) and initial conditions \(y(0)=0\) and \(y'(0)=1\).

Step-by-step solution

Find the first and second derivatives of y(x)

To find the first derivative, use the chain rule and the fact that the derivative of \(e^{ax}\) is \(ae^{ax}\) and the derivative of \(e^{-ax}\) is \(-ae^{-ax}\). $$ y'(x)=\frac{1}{4}(2e^{2 x}-(-2)e^{-2 x}) $$ Next, take the second derivative of y(x): $$ y''(x)=\frac{1}{4}(4e^{2 x}-(-4)e^{-2 x}) $$

Verify if y''(x)-4y(x)=0

Now we need to check if the given differential equation \(y''(x)-4y(x)=0\) holds true for our function y(x) and its second derivative. Substitute the values of y(x) and y''(x): $$ \frac{1}{4}(4e^{2 x}+4e^{-2 x})-4\left(\frac{1}{4}\left(e^{2 x}-e^{-2 x}\right)\right) $$ Simplify the equation: $$ (e^{2 x}+e^{-2 x})-(e^{2 x}-e^{-2 x}) $$ The terms cancel out, and we get: $$ 0=0 $$ Thus, the function satisfies the given differential equation.

Verify the initial conditions

Now let's test if the function y(x) satisfies the initial conditions y(0)=0 and y'(0)=1: Evaluate y(0): $$ y(0)=\frac{1}{4}\left(e^{2(0)}-e^{-2(0)}\right)=\frac{1}{4}(1-1)=0 $$ Evaluate y'(0): $$ y'(0)=\frac{1}{4}(2e^{2(0)}+2e^{-2(0)})=\frac{1}{4}(2+2)=1 $$ The function satisfies both initial conditions, y(0)=0 and y'(0)=1.

Conclusion

Given function \(y(x)=\frac{1}{4}\left(e^{2 x}-e^{-2 x}\right)\) is indeed a solution of the initial value problem with the differential equation \(y''(x) - 4y(x) = 0\) and initial conditions \(y(0)=0\), \(y'(0)=1\).

Key concepts

Initial Value Problem

An initial value problem in differential equations includes finding a function that satisfies a specific differential equation and fulfills certain conditions at a predefined point. To tackle this, we first need to know:

• The differential equation itself, such as the one in our example where the expression is given by \( y''(x) - 4y(x) = 0 \).
• The initial conditions, which specify the value of the function and possibly its derivatives at a certain point, like \( y(0) = 0 \) and \( y'(0) = 1 \) in this problem.

Having initial conditions is crucial because many different functions can solve a differential equation, but initial conditions help zero in on the one specific solution needed. In practical terms, it's like having a starting point and direction for a journey: without them, there’d just be too many possibilities to consider.

Solution Verification

Solution verification is a crucial step when dealing with differential equations, as it ensures that the proposed solution actually satisfies both the equation itself and the initial conditions stated. To verify a solution:

• First, substitute the function into the differential equation. For example, plugging our proposed solution \( y(x) = \frac{1}{4}(e^{2x} - e^{-2x}) \) into the equation \( y''(x) - 4y(x) = 0 \) to check if the left-hand side simplifies to zero.
• Second, ensure that the initial conditions hold by substituting them into the solution. In our example, evaluating \( y(0) \) and \( y'(0) \) showed us that both conditions \( y(0) = 0 \) and \( y'(0) = 1 \) were met.

By confirming both the equation and initial conditions, we could confidently assert that the given function is indeed the solution to the problem.

Chain Rule

The chain rule is an essential tool when finding derivatives of functions that involve compositions of functions. It's especially useful in calculus when dealing with exponential functions. Let's see how it applies here.
For a function composed of an inner function \( u(x) \) and an outer function \( y(u) \), the chain rule states that the derivative \( y'(x) \) is given by \( y'(u) \cdot u'(x) \). In our exercise:

• Consider finding \( y'(x) \) for \( y(x) = \frac{1}{4}(e^{2x} - e^{-2x}) \). Here, the chain rule helps when differentiating terms like \( e^{2x} \) where \( u(x) = 2x \) and \( y(u) = e^u \).
• The derivative \( e^{2x} \) requires multiplying the outer derivative \( e^{2x} \) by the derivative of the inner function, which is \( 2 \), so it becomes \( 2e^{2x} \).

Same logic applies to \( e^{-2x} \), with \(-2 \), completing the derivative calculation. Chain rule simplifies what could otherwise be a complex process of differentiation.

Exponential Functions

Exponential functions are particular types of mathematical functions with the form \( a^x \), where \( e^x \) (Euler's number, approximately 2.718) is a common base. These functions are known for their:

• Rapid growth or decay, exemplified by how they appear in natural phenomena involving growth processes.
• Unique properties, especially in calculus. Crucial identities include the fact that the derivative of \( e^{ax} \) is simply \( ae^{ax} \), making calculations straightforward.

In our example, the function \( y(x) = \frac{1}{4}(e^{2x} - e^{-2x}) \) required an understanding of these properties to derive and verify the solution through differentiation.
Knowing exponential functions helps in numerous fields like physics, engineering, and finance, where natural growth patterns are analyzed using calculus.