Problem 19 - Calculus

Question

Find the values of \(k\) for which \(y=x^{2}+k\) is a solution to the differential equation $$2 y-x y^{\prime}=10.$$

Step-by-Step Solution

Answer

The value of \(k\) is 5.

Step 1: Identify Components of Functions

The differential equation provided is \(2y - xy' = 10\) and the solution function is \(y = x^2 + k\). We need to find the derivative \(y'\) of this function.

Step 2: Calculate Derivative

Differentiate the function \(y = x^2 + k\) with respect to \(x\). The derivative \(y'\) is the derivative of \(x^2\) which equals \(2x\) and since \(k\) is a constant, its derivative is zero. Hence, \(y' = 2x\).

Step 3: Substitute Function and Derivative into Differential Equation

Substitute \(y = x^2 + k\) and \(y' = 2x\) into the differential equation \(2y - xy' = 10\), so we have: \[ 2(x^2 + k) - x(2x) = 10. \]

Step 4: Simplify Substitution

Simplify the equation from Step 3: \[ 2x^2 + 2k - 2x^2 = 10. \]

Step 5: Resolve the Equation

The \(2x^2\) terms cancel each other out, leading to the equation \[2k = 10.\] Divide both sides by 2 to solve for \(k\), resulting in \(k = 5\).

Key Concepts

Derivative

A derivative represents the rate of change of a function with respect to a variable. In simple terms, it tells us how a function changes as its input changes. For any function like \(y = x^2 + k\), the derivative with respect to \(x\) is found using basic differentiation rules.
The component \(x^2\) has a derivative of \(2x\) because the power rule states that the derivative of \(x^n\) is \(nx^{n-1}\). Since \(k\) is a constant, its derivative is zero because constants do not change and thus have no rate of change.
This leads us to the overall derivative \(y' = 2x\). This derivative is crucial as it is used in the verification of whether the function satisfies the given differential equation.

Solution Verification

Once you find the derivative, the next step is to verify if the solution satisfies the differential equation. Verification involves substituting both the function \(y\) and its derivative \(y'\) back into the equation provided.
For our exercise, substituting \(y = x^2 + k\) and \(y' = 2x\) into the equation \(2y - xy' = 10\), transforms it into the expression:

• \(2(x^2 + k) - x(2x) = 10\).

By simplifying this expression, you can check if the left-hand side equals the right-hand side of the equation given. If both sides match for specific values of \(k\), it confirms that \(y = x^2 + k\) is indeed a solution for those \(k\) values.
This step is about ensuring that all conditions of the differential equation are satisfied, which is essential for the correctness of the solution.

Algebraic Manipulation

Algebraic manipulation is a powerful tool used to solve and simplify equations. In this scenario, it's employed to isolate the variable \(k\) in the differential equation.
Starting with the equation \(2(x^2 + k) - x(2x) = 10\):

• First, distribute \(2\) across \(x^2 + k\), yielding \(2x^2 + 2k\).
• Next, distribute \(-x\) across \(2x\), resulting in \(- 2x^2\).
• Combining these, the expression becomes \(2x^2 + 2k - 2x^2 = 10\).

Notice how the \(2x^2\) terms cancel, simplifying the equation to \(2k = 10\). Divide both sides by \(2\) to solve for \(k\), yielding \(k=5\).
This process highlights the importance of careful arithmetic operations and simplifications in deriving the values of \(k\) that solve the differential equation.