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Chapter 1: Problem 2

Solve the ODE by integration. $$y^{\prime}=e^{-3 x}$$

Short Answer

Expert verified
\( y = -\frac{1}{3} e^{-3x} + C \) is the general solution.

Step by step solution

01

Identify the Integrable Form

The ordinary differential equation (ODE) given is \( y' = e^{-3x} \). This is a first-order ODE which can be solved by directly integrating the right side with respect to \( x \).
02

Integrate the Right Side

We integrate the function on the right side: \[\int e^{-3x} \, dx.\]To solve this integral, use the basic integration formula for exponentials. The integral becomes:\[-\frac{1}{3} e^{-3x} + C,\]where \( C \) is the constant of integration.
03

Write the General Solution

Since the left-hand side of the original equation was the derivative \( y' \), the function \( y \) itself is equal to the result of the integration:\[ y = -\frac{1}{3} e^{-3x} + C. \]

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus and serves as a powerful tool to solve first-order differential equations like our example. When we tackle the problem of finding a function whose derivative matches a given expression, we are essentially looking to "reverse" the process of differentiation. By integrating the given expression, we can uncover the original function with respect to the variable.
  • For the ODE given by \( y' = e^{-3x} \), integrating the right side \( e^{-3x} \) with respect to \( x \) allows us to find the function \( y \).
  • The integration process involves finding an antiderivative, which includes identifying any necessary constants that will balance the equation.
Integral calculus, therefore, helps us not just solve equations, but also understand how functions develop and change. When solving an integral that involves an expression of an exponential function, knowing integration rules can ease the process immensely.
Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They're essential in representing growth and decay scenarios in real-world situations.In our case, the differential equation contains the exponential function \( e^{-3x} \).
  • Exponential decay is a typical characteristic seen in functions where the exponent is a negative number.
  • Understanding how exponential functions behave under integration is crucial to derive accurate solutions.
Upon integration, exponential functions tend to retain their base form, but may acquire coefficients that arise from constants involved in the integration.
Constant of Integration
When you integrate a function, you'll often see the inclusion of an arbitrary constant, labeled as \( C \), called the constant of integration.
  • This is because integration is essentially the reverse process of differentiation, which, upon differentiation of a constant term, yields zero.
  • In the step-by-step solution provided for the exercise, the \( C \) ensures completeness of the solution by reflecting any constant that might have disappeared during differentiation. It represents the family of solutions that fit the original differentiable function.
Adding \( C \) to our solution takes into account all the possible vertical shifts of the curve that do not affect the slope or shape of \( y = -\frac{1}{3} e^{-3x} \). This concept underscores the idea that there's more than one function that can have the same derivative when constants are involved.

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Most popular questions from this chapter

(lntegration constant) Give a reason why in (4) you may choose the constant of integration in \(\int p d x\) to be zero.

Find all initial conditions such that \(\left(x^{2}-4 x\right) y^{\prime}=(2 x-4) y\) has no solution, precisely one solution, and more than one solution.

Verify that \(y\) is a solution of the ODE Determine from \(y\) the particular solution satisfying the given initial condition. Sketch or graph this solution. $$y^{\prime}=0.5 y, \quad y=c e^{0.5 x}, \quad y(2)=2$$

Test for exactness. If exact, solve, If not, use an integrating factor as given or find it by inspection or from the theorems in the text. Also, if an initial condition is given, determine the corresponding particular solution. $$-\pi \sin \pi x \sinh y d x+\cos x \cosh y d y=0$$

Find a general solution. Show the steps of derivation. Check your answer by substitution. $$y^{\prime} \sin \pi x=y \cos x$$
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