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Chapter 8: Problem 42

Verify that the given function is a solution of the differential equation. $$ y^{\prime \prime}-2 y^{\prime}+y=0 ; y=-2 e^{x}+x e^{x} $$

Short Answer

Expert verified
The function y = -2e^{x} + xe^{x} is a solution of the differential equation y'' - 2y' + y = 0.

Step by step solution

01

Find the First Derivative

To verify if the function is a solution to the differential equation, first find the first derivative of the given function. The function is: y = -2e^{x} + xe^{x}Using the product rule for differentiation, compute: y' = d/dx(-2e^{x}) + d/dx(xe^{x})y' = -2e^{x} + (1)e^{x} + x(e^{x})y' = -2e^{x} + e^{x} + xe^{x}y' = -e^{x} + xe^{x}
02

Find the Second Derivative

Now find the second derivative y'' = d/dx(-e^{x} + xe^{x})Using the product rule again:y'' = d/dx(-e^{x}) + d/dx(xe^{x})y'' = -e^{x} + (1)e^{x} + x(e^{x})y'' = -e^{x} + e^{x} + xe^{x}y'' = -e^{x} + e^{x} + xe^{x}y'' = xe^{x}
03

Substitute into the Differential Equation

Substitute the function and its derivatives into the differential equation y'' - 2y' + y = 0We have y'' = xe^{x}, y' = -e^{x} + xe^{x}Substitute these into the equation:y'' - 2y' + y = xe^{x} - 2(-e^{x} + xe^{x}) + (-2e^{x} + xe^{x})Simplifying:= xe^{x} - 2(-e^{x}) - 2(xe^{x}) -2e^{x} + xe^{x}= xe^{x} + 2e^{x} - 2xe^{x} - 2e^{x} + xe^{x}= xe^{x} - 2xe^{x} + xe^{x}= 0
04

Conclusion

Since the left side of the differential equation equals zero (0), the given function satisfies the differential equation. Therefore, the function y = -2e^{x} + xe^{x}is a solution to the differential equation y'' - 2y' + y = 0.

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Key Concepts

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

First Derivative
Understanding the first derivative is crucial in verifying solutions to differential equations. A derivative represents how a function changes as its input changes. Essentially, it gives the slope of the function at any point. In our problem, we start with the function: \[ y = -2e^{x} + xe^{x} \] To find the first derivative, apply the basic derivatives of exponential functions and the **Product Rule**.
Remember, the product rule states: \( (uv)' = u'v + uv' \).

For our function: \[ y' = d/dx(-2e^{x}) + d/dx(xe^{x}) \] The calculations follow:
  • Derivative of \( -2e^{x} \): \[ -2e^{x} \]
  • Derivative of \( x e^{x} \): First apply product rule where \[ u = x \] and \[ v = e^{x} \]. Therefore, \[ u' = 1 \] and \[ v' = e^{x} \]
So,
  • \[ d/dx(xe^{x}) = (1)e^{x} + x(e^{x}) = e^{x} + x(e^{x}) \]
Putting it together: \[ y' = -2e^{x} + e^{x} + xe^{x} = -e^{x} + xe^{x} \]
Second Derivative
The second derivative gives us information about the curvature of the function's graph - whether it curves upwards or downwards. In verification problems, we often need this second derivative. We move forward by differentiating the first derivative: \[ y'' = d/dx(-e^{x} + xe^{x}) \] Using the same principles, the product rule applies again.
Let's break it down:
  • Derivative of \( -e^{x} \): \[ -e^{x} \]
  • Using the product rule on \( x e^{x} \):
    \[ u = x \] and \[ v = e^{x} \], so \[ u' = 1 \] and \[ v' = e^{x} \]
    Hence, \[ d/dx(x e^{x}) = (1)e^{x} + x(e^{x}) = e^{x} + x(e^{x}) \]
Therefore,
\[ y'' = -e^{x} + e^{x} + x(e^{x}) \ = -e^{x} + e^{x} + x(e^{x}) \ = x e^{x} \]
Product Rule
The product rule is a vital tool for finding derivatives of products of two functions.
To restate the rule: \( (uv)' = u'v + uv' \).
Let's apply it step by step for practice:
Suppose we have \[ u(x) \] and \[ v(x) \]. Then their product is \[ u(x) \times v(x) \]. According to the product rule:
\( f'(x) = u'(x) \times v(x) + u(x) \times v'(x) \).
  • Take \[ y = x e^{x} \]. Here, \[ y' = (1) e^{x} + x (e^{x}) = e^{x} + x e^{x} \]
For another example, think of our previous derivative problem:
We started with \[ y = -2 e^{x} + x e^{x} \]. By breaking it into parts,
  • The derivative of \[ -2 e^{x} \] is \[ -2 e^{x} \]
  • For the product \[ x \times e^{x} \], we use the product rule.
    Thus, \[ (x e^{x})' = e^{x} + x e^{x} \]
This leads to combining and simplifying to get our first derivative.
Grasping the product rule thoroughly helps ease more complex derivative calculations.

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Most popular questions from this chapter

Solve the initial-value problem. State an interval on which the solution exists. $$ y^{\prime}+4 t y=t ; y(0)=3 $$

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Solve the initial-value problem. $$ y^{\prime}=x e^{x} ; y(0)=2 $$
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