Problem 11 - Calculus and its applications

Question

Let \(y^{\prime}+4 y=0\) a) Show that \(y=e^{-4 x}\) is a solution of this differential equation. b) Show that \(y=C e^{-4 x}\) is a solution, where \(C\) is a constant.

Step-by-Step Solution

Answer

Both functions, \( y = e^{-4x} \) and \( y = Ce^{-4x} \), are solutions to the differential equation.

Step 1: Differentiate the given solution

Given the potential solution \(y = e^{-4x}\), we need to find \(y'\). To do this, differentiate \(y = e^{-4x}\) with respect to \(x\). Using the chain rule: \(y' = -4e^{-4x}\).

Step 2: Substitute into the differential equation

Substitute \(y = e^{-4x}\) and \(y' = -4e^{-4x}\) into the differential equation \(y' + 4y = 0\). This gives:\[-4e^{-4x} + 4e^{-4x} = 0.\]

Step 3: Simplify the equation

The equation simplifies to:\[0 = 0,\]confirming that the equation holds true. This verifies that \(y = e^{-4x}\) is indeed a solution to the differential equation \(y' + 4y = 0\).

Step 4: Differentiate the general solution with constant

Consider the general solution \(y = Ce^{-4x}\), where \(C\) is a constant. Differentiate it with respect to \(x\) to get:\(y' = C(-4)e^{-4x} = -4Ce^{-4x}.\)

Step 5: Substitute the general solution into the differential equation

Substitute \(y = Ce^{-4x}\) and \(y' = -4Ce^{-4x}\) into the differential equation:\(-4Ce^{-4x} + 4Ce^{-4x} = 0.\)

Step 6: Verify the solution

The equation simplifies to:\[0 = 0,\]which confirms that \(y = Ce^{-4x}\) is a valid solution to the differential equation \(y' + 4y = 0\).

Key Concepts

Solution Verification

Solution verification involves checking whether a proposed function satisfies a given differential equation. To verify a solution, you'll often follow these steps:

• Differentiate the function: Calculate the derivative of the proposed solution. For instance, when given a function like \(y = e^{-4x}\), you find \(y' = -4e^{-4x}\) using differentiation rules.
• Substitute into the differential equation: Insert both the function and its derivative back into the original differential equation. For our exercise, the equation is \(y' + 4y = 0\).
• Check for equalities: Ensure that after substitution, the left-hand side and the right-hand side of the equation match. For instance, \(-4e^{-4x} + 4e^{-4x} = 0\) confirms the solution because it simplifies to \(0 = 0\).

By following these steps, you can systematically verify that a function is indeed a solution to the given differential equation.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. It's particularly handy when dealing with exponentials involving complex exponents.When differentiating a function like \(y = e^{-4x}\), the chain rule helps by acknowledging the inner function, which in this instance is \(-4x\). The chain rule states:\[\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)\]For \(y = e^{-4x}\):

• Identify the outer function: For \(e^{-4x}\), the outer function is the exponential function \(e^u\), where \(u = -4x\).
• Differentiate the outer function: The derivative of \(e^u\) with respect to \(u\) is \(e^u\).
• Multiply by the derivative of the inner function: Since \(u = -4x\), its derivative is \(-4\). Thus, \(d/dx[e^{-4x}] = (-4)e^{-4x}\).

Using the chain rule effectively allows us to quickly find that \(y' = -4e^{-4x}\), crucially aiding in both understanding and solving differential equations.

Exponential Functions

Exponential functions are pivotal in calculus, especially when solving differential equations where growth and decay are modeled. The form \(y = e^{kx}\), where \(k\) is a constant, is a typical example.Key properties of exponential functions include:

• Continuous Growth/Decay: Depending on the sign of \(k\), the function depicts exponential growth (\(k > 0\)) or decay (\(k < 0\)). In \(y = e^{-4x}\), the negative sign indicates a steadily decreasing function as \(x\) increases.
• Differentiation: The derivative of \(e^{kx}\) remains an exponential function, ##: \(\frac{d}{dx} [e^{kx}] = ke^{kx}\). This property is crucial for differential equations as it enables simple substitution.
• Smooth Curve and Domain: Exponential functions are always smooth and defined for all real numbers, which makes them easy to work with in many mathematical contexts.

Understanding these properties allows better application of exponential functions in solving differential equations, making them an essential tool for analysts and engineers alike.