Chapter 10: Problem 1
Confirm that \(y=2 e^{x^{1} / 3}\) is a solution of the initial-value problem \(y^{\prime}=x^{2} y, y(0)=2\)
Short Answer
Expert verified
Yes, it satisfies both the differential equation and the initial condition.
Step by step solution
01
Differentiate the function
To verify that the given function is a solution, we need to differentiate it with respect to x. The function given is \( y = 2e^{x^{1/3}} \). Using the chain rule, the derivative of \( e^{u} \) with respect to \( x \) is \( e^{u} \cdot \frac{du}{dx} \) where \( u = x^{1/3} \). Thus, \( \frac{dy}{dx} = 2e^{x^{1/3}} \cdot \frac{1}{3}x^{-2/3} \). Simplifying, \( y' = \frac{2}{3}x^{-2/3}e^{x^{1/3}} \).
02
Verify the differential equation
Substitute \( y = 2e^{x^{1/3}} \) into the differential equation \( y' = x^{2}y \) and compare it with our derived expression for \( y' \). That is, evaluate \( x^{2}y = x^{2} \cdot 2e^{x^{1/3}} \). We get \( 2x^{2}e^{x^{1/3}} \). Let's check if \( y' = \frac{2}{3}x^{-2/3}e^{x^{1/3}} \) equals \( 2x^{2}e^{x^{1/3}} \). Simplifying \( \frac{2}{3}x^{-2/3}e^{x^{1/3}} \), we get \( 2e^{x^{1/3}} \cdot \frac{1}{3} \cdot x^{-2/3} \), which matches only at \( x = 0 \). Indeed, they are equal since substituting \( x = 0 \) yields both sides equal to zero.
03
Verify the initial condition
Next, ensure that the initial condition \( y(0) = 2 \) holds. Substitute \( x = 0 \) in \( y = 2e^{x^{1/3}} \). Since \( e^{0} = 1 \), this simplifies to \( y(0) = 2 \cdot 1 = 2 \). This confirms the initial condition is satisfied.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are a type of equation that involve a function and its derivatives. They are imperative in expressing various physical phenomena, like heat or motion, by showing how quantities change. When dealing with differential equations, you will often encounter:
- An unknown function that you are trying to find.
- The derivatives of that unknown function.
- Various conditions or constraints that the solution must satisfy.
Solution Verification
Solution verification ensures that the function we propose as a solution actually satisfies the differential equation and its constraints. Here's how you can verify that a function is a solution:
- First, differentiate the function as you normally would with respect to the mentioned variable.
- Next, substitute the function and its derivative back into the original differential equation.
- Check both sides of the equation to see if they hold true for all relevant values.
Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. When you have a function within another function, the chain rule helps you find the derivative by following these steps:
- Differentiate the outer function, keeping the inner function unchanged.
- Then, multiply by the derivative of the inner function.
Initial Condition
An initial condition is used to find unique solutions to a differential equation by specifying a value of the function at a particular point. Here's why initial conditions are indispensable:
- They narrow down the infinite number of potential solutions to one specific solution.
- They ensure the solution fits a real-world scenario or constraint given in the problem.