Problem 52 Show that (a) \(y=x e^{-x}\) s... [FREE SOLUTION]

Chapter 4: Problem 52

Show that (a) \(y=x e^{-x}\) satisfies the equation \(x y^{\prime}=(1-x) y\) (b) \(y=x e^{-x^{2} / 2}\) satisfies the equation \(x y^{\prime}=\left(1-x^{2}\right) y\)

Short Answer

Expert verified

Both functions satisfy their respective equations by substituting the derived \( y' \) and verifying equality.

Step by step solution

Differentiate y for (a)

The function given is \( y = x e^{-x} \). To find \( y' \), we apply the product rule: \( (uv)' = u'v + uv' \). Let \( u = x \) and \( v = e^{-x} \). Thus, \( u' = 1 \) and \( v' = -e^{-x} \). The derivative is:\[y' = (x)' e^{-x} + x (e^{-x})' = e^{-x} + x(-e^{-x}) = e^{-x} - xe^{-x}.\]

Substitute y and y' in (a)

Substitute \( y = x e^{-x} \) and \( y' = e^{-x} - xe^{-x} \) into the equation \( xy' = (1-x)y \):\[xy' = x(e^{-x} - xe^{-x}) = xe^{-x} - x^2 e^{-x}\]\[(1-x)y = (1-x)(x e^{-x}) = x e^{-x} - x^2 e^{-x}\]These are identical, so the given \( y \) satisfies the equation.

Differentiate y for (b)

The function is \( y = x e^{-x^2/2} \). Use the product rule again:Let \( u = x \) and \( v = e^{-x^2/2} \). Then \( u' = 1 \) and \( v' = -x e^{-x^2/2} \).\[y' = (x)' e^{-x^2/2} + x(-x e^{-x^2/2}) = e^{-x^2/2} - x^2 e^{-x^2/2}.\]

Substitute y and y' in (b)

Substitute \( y = x e^{-x^2/2} \) and \( y' = e^{-x^2/2} - x^2 e^{-x^2/2} \) into the equation \( xy' = (1-x^2)y \):\[xy' = x(e^{-x^2/2} - x^2 e^{-x^2/2}) = xe^{-x^2/2} - x^3 e^{-x^2/2}\]\[(1-x^2)y = (1-x^2)(x e^{-x^2/2}) = x e^{-x^2/2} - x^3 e^{-x^2/2}\]Again, the expressions match, verifying the solution.

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Key Concepts

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

Product Rule in Calculus

The product rule is a fundamental tool in calculus, mainly used to differentiate functions that are products of two or more simpler functions. If you have a function defined as the product of two functions, say, \( u(x) \) and \( v(x) \), then the product rule gives us a way to find the derivative without having to expand the product first. The rule is simple and elegant:

If \( y = u(x) \, v(x) \), then the derivative \( y' = u'(x) \, v(x) + u(x) \, v'(x) \).
- This means you differentiate \( u(x) \) and \( v(x) \) separately and combine the results as shown.

For example, in our exercise, we had to differentiate \( y = x \, e^{-x} \). By applying the product rule:

Let \( u = x \), so \( u' = 1 \).
Let \( v = e^{-x} \), so \( v' = -e^{-x} \).

Now, substitute:

\( y' = 1 \, e^{-x} + x(-e^{-x}) = e^{-x} - x \, e^{-x} \).

The product rule simplifies what could otherwise be a cumbersome process of multiplying out terms before differentiating.

Verification of Solutions in Differential Equations

When dealing with differential equations, verifying a solution means showing that a purported solution actually satisfies the original differential equation. This is typically done by substituting the given solution into the equation.For example, to verify that \( y = x \, e^{-x} \) is a solution to \( x y' = (1-x)y \), follow these steps:

First, we find the derivative \( y' \), using the product rule. Here, it was calculated as \( e^{-x} - x \, e^{-x} \).
Then substitute both \( y \) and \( y' \) back into the differential equation.
The equation becomes \( x(e^{-x} - x \, e^{-x}) = (1-x)(x \, e^{-x}) \).
- Simplifying both sides, they equate to \( x \, e^{-x} - x^2 \, e^{-x} \).
- Both sides match, hence the solution is verified.

By verifying both sides of an equation separately, you confirm that your proposed function behaves as needed under the given conditions, which is critical in solving differential equations correctly.

Understanding Exponential Functions

Exponential functions are a type of mathematical function that involve the constant \( e \), where \( e \approx 2.71828 \). These functions are characterized by a variable exponent and are widely used in various scientific fields.Key features of exponential functions:

The function \( e^x \) grows rapidly as \( x \) increases, and approaches zero as \( x \) decreases.
- This property makes them useful in modeling growth and decay processes, such as population growth or radioactive decay.
In calculus, exponential functions have unique properties:
- The derivative of \( e^x \) is itself \( e^x \), providing simplicity in differentiation tasks.
- For a decay function like \( e^{-x} \), its derivative is \(-e^{-x} \), indicating a decrease in value with increasing \( x \).

Applying these characteristics in exercises, such as differentiating and verifying solutions, allows students to appreciate the powerful simplicity and applicability of exponential functions across various domains of mathematics and science.

91影视

Show that (a) \(y=x e^{-x}\) satisfies the equation \(x y^{\prime}=(1-x) y\) (b) \(y=x e^{-x^{2} / 2}\) satisfies the equation \(x y^{\prime}=\left(1-x^{2}\right) y\)

Short Answer

Step by step solution

Differentiate y for (a)

Substitute y and y' in (a)

Differentiate y for (b)

Substitute y and y' in (b)

Key Concepts

Product Rule in Calculus

Verification of Solutions in Differential Equations

Understanding Exponential Functions

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