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Prove that \(y\) is a solution of the differential equation. $$ y^{\prime}+3 y=0 ; \quad y=C e^{-3 x} $$

Short Answer

Expert verified
Yes, \( y = Ce^{-3x} \) is a solution of \( y' + 3y = 0 \).

Step by step solution

01

Differentiate y with respect to x

Given function is \( y = Ce^{-3x} \). Find the derivative of \( y \) with respect to \( x \). Therefore, using the chain rule, we have: \( y' = \frac{d}{dx}(Ce^{-3x}) \). The derivative is \( y' = -3Ce^{-3x} \).
02

Substitute y and y' in the Differential Equation

The differential equation is \( y' + 3y = 0 \). Substitute \( y' = -3Ce^{-3x} \) and \( y = Ce^{-3x} \) into this equation. It becomes: \( -3Ce^{-3x} + 3(Ce^{-3x}) = 0 \).
03

Simplify the Expression

The equation \( -3Ce^{-3x} + 3Ce^{-3x} = 0 \) can be simplified. Adding these together gives \( 0 = 0 \), showcasing that the expression simplifies correctly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Chain Rule
The chain rule is a fundamental concept in calculus used to find the derivative of composite functions. When dealing with a function like \( y = Ce^{-3x} \), you notice that \( e^{-3x} \) is a composite function. It contains an outer exponential function and an inner linear function \(-3x\).

The chain rule states that to take the derivative of a composite function, you must take the derivative of the outer function and multiply it by the derivative of the inner function. In our example:
  • The outer function is \( e^u \) where \( u = -3x \); its derivative is \( e^u \).
  • The inner function \( -3x \) has a derivative of \( -3 \).
Now, to differentiate \( Ce^{-3x} \), apply the chain rule. First, pull down the constant \( C \), then multiply by the derivative of the exponential, \( e^{-3x} \), followed by the derivative of the exponent \(-3\):

\[ y' = C(-3)e^{-3x} = -3Ce^{-3x} \] This result is crucial for solution verification later.
Ensuring Solution Verification
Solution verification involves checking whether a proposed solution satisfies a given differential equation. This means ensuring that when substituting into the equation, both sides match or simplify to the same statement, usually something true like \( 0 = 0 \).

Let's recap what we've got:
  • Our original differential equation is \( y' + 3y = 0 \).
  • We calculated \( y' = -3Ce^{-3x} \)._
  • Our given solution is \( y = Ce^{-3x} \).
Substitute these into the equation to verify the solution.

The left-hand side becomes:

\[ y' + 3y = -3Ce^{-3x} + 3(Ce^{-3x}) \]

On simplifying, all terms cancel out, resulting in \( 0 \). So the left side equals the right side (which is also \( 0 \)), confirming the proposed solution indeed satisfies the differential equation.
Exploring the Exponential Function
The exponential function \( e^x \) is a key component in this differential equation. Its unique properties make it essential in calculus and beyond. For the given solution \( y = Ce^{-3x} \), \( C \) represents a constant, and \( e^{-3x} \) models an exponential decay process.

There are some important features of exponential functions:
  • Exponential functions are incredibly versatile, appearing in biology, economics, and physics.
  • The base \( e \) is approximately equal to 2.718, known as Euler's number, and it has a unique property where the derivative of \( e^x \) is itself \( e^x \).
In our equation, \( e^{-3x} \) signifies that as \( x \) increases, \( y \) decreases exponentially. Such behavior is typical in decay processes like radioactive decay or cooling objects. Understanding these dynamics gives insights into how solutions relate to real-world situations.

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