Question

Verify that the following functions are solutions to the given differential equation.\(y=3-x+x \ln x\) solves \(y^{\prime}=\ln x\)

Short answer

Yes, the function is a solution to the differential equation.

Step-by-step solution

Differentiate the given function

We need to differentiate the function \( y = 3 - x + x \ln x \) with respect to \( x \). To do this, apply the product rule to the term \( x \ln x \).\m \( \frac{d}{dx}(x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \). Thus, \m \( y' = \frac{d}{dx}(3) - \frac{d}{dx}(x) + \frac{d}{dx}(x \ln x) \), we have \( y' = 0 - 1 + (\ln x + 1) = \ln x \).

Verify the differentiated function matches the original differential equation

The derivative we found is \( y' = \ln x \). The given differential equation is also \( y' = \ln x \). Since both are equal, \( y = 3 - x + x \ln x \) is a solution to the differential equation \( y' = \ln x \).

Key concepts

Product Rule

When differentiating functions, particularly those in the format of one function multiplied by another, the product rule is essential. It's used when you encounter expressions like \( x \ln x \), where both \( x \) and \( \ln x \) are functions of \( x \). The product rule in calculus states that if you have two functions, \( u(x) \) and \( v(x) \), then the derivative of their product is given by:

• \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

In our exercise, given \( y = 3 - x + x \ln x \), we applied the product rule to the term \( x \ln x \). Here, \( u(x) = x \) and \( v(x) = \ln x \). Differentiating, we find:

• \( u'(x) = 1 \)
• \( v'(x) = \frac{1}{x} \)

So, \( \frac{d}{dx}(x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \). This indicates how the product rule helps break down complex differentiations into manageable steps.

Differentiation

Differentiation is a fundamental operation in calculus that involves finding the rate at which a function changes at any point. In simpler terms, it seeks to find the slope of the curve of a function. For the function \( y = 3 - x + x \ln x \), the task is to differentiate each component with respect to \( x \):

• The derivative of a constant, \( 3 \), is \( 0 \).
• The derivative of \( -x \) is \( -1 \).
• For the term \( x \ln x \), we utilize the product rule, resulting in \( \ln x + 1 \).

Hence, the derivative \( y' \) becomes \( 0 - 1 + (\ln x + 1) \). Simplifying, we get \( y' = \ln x \). This process demonstrates how differentiation is systematically applied, term by term, to arrive at the derivative of the entire function.

Verify Solutions

The final step in solving many differential equations is verifying that a proposed solution correctly satisfies the equation. This involves substituting the differentiated result back into the original differential equation and confirming that both sides are equal. In our case, the given differential equation is \( y' = \ln x \).

• Firstly, we differentiated the function \( y = 3 - x + x \ln x \) and found \( y' = \ln x \).
• Next, we compared this to the given differential equation. Since both expressions for \( y' \) are identical, we confirm that the function satisfies the differential equation.

This verification ensures that the proposed function is indeed a solution, affirming the accuracy of the differentiation process applied. Ensuring both sides of the equation match is crucial in confirming the correctness of solutions in calculus.