Question

Determining a Solution In Exercises \(21-28\) , determine whether the function is a solution of the differential equation \(x y^{\prime}-2 y=x^{3} e^{x} .\) $$ y=x^{2}\left(2+e^{x}\right) $$

Short answer

The given function is indeed a solution if both sides of the equation are equal after substituting the differentiated function and simplifying the result.

Step-by-step solution

Differentiate the Function

Begin by differentiating the given function with respect to \(x\). The function is \(y=x^{2}\left(2+e^{x}\right)\). By applying the product and the chain rule, the derivative, \(y'\), can be found.

Substitute in the Equation

Next, substitute the calculated derivative, \(y'\), and \(y\) value into the differential equation \(x y^{\prime}-2 y=x^{3} e^{x}\). This operation will generate an expression.

Simplify the Resulting Expression

After the substitution, simplify the expression obtained to verify if both sides of the equation are equal. If the differential equation simplifies and holds true, then the initial function is indeed a solution to the differential equation.

Key concepts

Product Rule

When dealing with functions composed of two or more factors that are dependent on the same variable, the product rule is a handy tool. The product rule states: if you have a function defined as the product of two functions, say \( f(x) = g(x) \cdot h(x) \), the derivative \( f'(x) \) is calculated as:

• \( f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x) \)

This means you differentiate the first function, leave the second as is, then leave the first function and differentiate the second, and finally add the results.
In our case, the function \( y = x^2 (2+e^x) \) needs this rule. Apply it by letting \( g(x) = x^2 \) and \( h(x) = 2+e^x \). Differentiate each:

• \( g'(x) = 2x \)
• \( h'(x) = e^x \)

Plug these into the product rule formula to find \( y' \), the derivative of \( y \), which is crucial in testing if \( y \) is a solution of a given differential equation.

Chain Rule

The chain rule is essential when a function is nested inside another function. It's a method used to differentiate composite functions, those that consist of one function inside another function. In simpler terms, if we have a composition \( u(v(x)) \), the derivative is:

• \( \frac{du}{dv} \cdot \frac{dv}{dx} \)

This rule is crucial when differentiating more complex expressions like \( e^x \) within our function \( y = x^2 (2+e^x) \).
In applying the chain rule to \( h(x) = 2 + e^x \), we note that the derivative of \( e^x \) with respect to \( x \) is simply \( e^x \).
Combining this with other derivatives calculated using the product rule ensures the full differentiation of \( y \). Correct usage of the chain rule ensures no steps are skipped,

Differentiation

Differentiation is the process of finding the derivative of a function, revealing its gradient or slope at any point. It assesses how the function changes relative to changes in its input. In the exercise, differentiating \( y = x^2 (2 + e^x) \) is key to determining if it satisfies the given differential equation.
Follow these steps to perform differentiation:

• Identify all distinct components of the function.
• Apply derivative rules - primarily the product and chain rules where necessary.
• Solve mathematically to find \( y' \).

Through differentiation, a picture of how our function behaves over its domain is established.
This insight enables hypothesis testing against differential equations. What remains is to check if the differentiated expression slotting into the equation balances both sides, proving or disproving \( y \) as a solution.

Solution Verification

Solution verification is the final step to determine whether the differentiated expression \( y' \) truly satisfies the given differential equation.
To verify, substitute the obtained \( y' \) and \( y \) back into the equation \( x y^{\prime} - 2y = x^3 e^x \). This helps form a new expression.

• Simplify this expression carefully.
• Compare both sides of the equation.

If both sides are equal after simplification, \( y \) is indeed a solution to the differential equation. This checks the work done in differentiation and rules application. If they do not match, re-check the differentiation steps and rule applications.
Solution verification confirms the chain of calculations and logical steps utilized, ensuring full understanding of the equation's dynamics and correctness of the function as a solution.