Problem 4 - Calculus

Question

Show that \(y(x)=A e^{\lambda x}\) is a solution to the equation \(y^{\prime}=\lambda y\) for any value of \(A.\)

Step-by-Step Solution

Answer

The function \(y(x)=A e^{\lambda x}\) satisfies the differential equation \(y' = \lambda y\) for any \(A\).

Step 1: Differentiate the function

First, we need to differentiate the function \(y(x) = A e^{\lambda x}\) with respect to \(x\). The derivative of \(e^{ax}\) with respect to \(x\) is \(a e^{ax}\). In this case, differentiating \(y(x)\), we get:\[ y'(x) = \frac{d}{dx}(A e^{\lambda x}) = A \lambda e^{\lambda x}. \]

Step 2: Substitute into the differential equation

Substitute \(y'(x)\) and \(y(x)\) back into the differential equation \(y' = \lambda y\). We identified in Step 1 that:\[ y'(x) = A \lambda e^{\lambda x}. \]For \(y(x) = A e^{\lambda x}\):\[ \lambda y(x) = \lambda (A e^{\lambda x}) = A \lambda e^{\lambda x}. \]

Step 3: Verify the equality

Using the expressions from Step 2, we equate\[ y'(x) = A \lambda e^{\lambda x} \]and\[ \lambda y(x) = A \lambda e^{\lambda x}. \]Since both sides of the equation \(y' = \lambda y\) are equal, we have shown that the function \(y(x) = A e^{\lambda x}\) satisfies the differential equation for any value of \(A\).

Key Concepts

Exponential Functions

Exponential functions are a fundamental part of mathematics and appear frequently in differential equations. They take the form \( f(x) = a e^{bx} \), where \( e \) is the base of natural logarithms, approximately equal to 2.71828. The variable \( x \) represents the exponent, and \( a \) and \( b \) are constants that define the specific behavior of the function. This type of function is important because it models processes that grow or decay at a constant rate, such as population growth or radioactive decay.

• Exponential functions grow or decay exponentially, which means their rate of change is proportional to their current value.
• The parameter \( b \) affects the rate of growth or decay. If \( b > 0 \), the function models exponential growth. Conversely, if \( b < 0 \), it models exponential decay.

Understanding these properties helps in solving differential equations, as they often require identifying solutions such as \( y(x) = A e^{x} \), which naturally satisfy many common equations. This forms the basis for their widespread application in various fields like physics, finance, and biology.

Derivatives

Differentiation is a critical concept in calculus, allowing us to determine how a function changes. For an exponential function like \( y(x) = A e^{x} \), its derivative \( y'(x) \) gives vital information about the rate of change, which is particularly useful in differential equations.

• To differentiate an exponential function \( e^{x} \), the rule \( \frac{d}{dx}(e^{x}) = b e^{x} \) applies. This means multiplying the original function by the coefficient of \( x \) in the exponent.
• In our exercise, differentiating \( y(x) = A e^{x} \) with respect to \( x \) yields \( y'(x) = A \lambda e^{\lambda x} \).

This result highlights how the structure of derivatives of exponential functions repeats the original function, only scaled by the factor in the exponent. As a result, they are exceptionally convenient in solving problems where proportional growth or decay is involved, as seen in the given differential equation \( y' = \lambda y \).

Solution Verification

Solution verification involves confirming whether a proposed solution truly satisfies a given problem. For differential equations, it's crucial to check if the function not only appears correct but also fits mathematically.
In the exercise, the verification entails substituting the arrived derivative and the original function back into the differential equation \( y' = \lambda y \). If both sides of the equation are equal after substitution, then \( y(x) = A e^{\lambda x} \) is indeed a solution for all values of \( A \).

• We take \( y'(x) = A \lambda e^{\lambda x} \) from differentiation. This matches \( \lambda y(x) = \lambda(A e^{\lambda x}) \), confirming the correctness of the solution.
• This step affirms that the proposed solution is valid for the specific structure of the differential equation, emphasizing the universality of exponential functions in such contexts.

Practicing solution verification strengthens problem-solving abilities, ensuring not only correctness but also deep comprehension of how the solution fits into the larger mathematical framework.