Problem 1 - Applied Calculus

Question

Decide whether or not each of the following is a solution to the differential equation \(x y^{\prime}-2 y=0\) (a) \(y=x^{2}\) (b) \(y=x^{3}\)

Step-by-Step Solution

Answer

\(y = x^2\) is a solution; \(y = x^3\) is not a solution.

Step 1: Differentiate the given solution

For each function provided, differentiate it with respect to x. (a) For the function \(y = x^2\), the derivative \(y' = \frac{d}{dx}(x^2) = 2x\).(b) For the function \(y = x^3\), the derivative \(y' = \frac{d}{dx}(x^3) = 3x^2\).

Step 2: Substitute into the differential equation

Substitute \(y\) and \(y'\) into the differential equation \(x y' - 2y = 0\).(a) For \(y = x^2\), substitute \(y = x^2\) and \(y' = 2x\) into the equation:\[x(2x) - 2(x^2) = 2x^2 - 2x^2 = 0\](b) For \(y = x^3\), substitute \(y = x^3\) and \(y' = 3x^2\) into the equation:\[x(3x^2) - 2(x^3) = 3x^3 - 2x^3 = x^3\]

Step 3: Evaluate the results

Check if each substitution satisfies the equation.(a) For \(y = x^2\), the result simplifies to \(0 = 0\), which satisfies the equation, indicating \(y = x^2\) is a solution.(b) For \(y = x^3\), the result is \(x^3 = 0\), which does not hold for all \(x\), indicating \(y = x^3\) is not a solution.

Key Concepts

Solution Verification

When working with differential equations, we often have potential solutions that need to be checked. This process is known as solution verification. It involves assessing whether a given function satisfies the original differential equation.
Here, the equation is linear in terms of the derivative, which is often a common type in calculus problems:

• First, take the derivative of the proposed solution.
• Next, substitute the function and its derivative back into the differential equation.
• Finally, simplify to check if both sides of the equation are equal.

This method ensures that the particular function is indeed a solution. Verification helps confirm your work and is essential especially when the stability of solutions or boundary conditions are involved.

Differentiation

Differentiation is a fundamental mathematical tool that's essential when dealing with differential equations. At its core, differentiation involves finding the derivative of a function, which essentially measures how the function changes at any point.
For the given problem, differentiation involves:

• Checking whether the function’s derivative can be substituted back into the equation.
• Using power rules, like for polynomials, which state that for any function of the form \(y = x^n\), the derivative is \(y' = nx^{n-1}\).

The step-by-step differentiation of the proposed solutions reveals their derivatives as illustrated in the solved example:- For \(y = x^2\), the derivative is \(2x\).- For \(y = x^3\), the derivative is \(3x^2\).Using these derivatives, we further validate whether these forms satisfy the differential equation given. Differentiation is thus pivotal in solving and verifying solutions of differential equations.

Substitution Method

The substitution method is often employed after differentiation to verify solutions of differential equations. This method works by plugging back the original function and its derivative into the differential equation to see if it satisfies the equation.
In our problem:

• For \(y = x^2\): the derivative \(2x\) and the function can be substituted back into the equation \(xy' - 2y = 0\). Simplifying, we find both sides equal, confirming this as a valid solution.
• For \(y = x^3\): after substitution, the results yield a mismatch, showing that the equation does not hold true for all \(x\).

Substitution plays a crucial role in verification and provides a straightforward path to confirm if solutions fit differential equations adequately. This method highlights how both the function and its derivative interact within the context of the equation itself.