Problem 2 - A First Course in Differential Equations with Modeling Applications

Question

Verify that \(y_{1}\) and \(y_{2}\) are solutions of the given differential equation but that \(y=c_{1} y_{1}+c_{2} y_{2}\) is, in general, not a solution. $$y y^{\prime \prime}=\frac{1}{2}\left(y^{\prime}\right)^{2} ; \quad y_{1}=1, y_{2}=x^{2}$$

Step-by-Step Solution

Answer

Both \(y_1=1\) and \(y_2=x^2\) are solutions, but \(y = c_1 y_1 + c_2 y_2\) is not a solution.

Step 1: Differentiate Solutions

First, we need to find the derivatives of the given solutions. For \(y_1 = 1\): \(y_1' = 0\) and \(y_1'' = 0\). For \(y_2 = x^2\): \(y_2' = 2x\) and \(y_2'' = 2\).

Step 2: Substitute into the Differential Equation for \(y_1\)

Substitute \(y_1, y_1', y_1''\) into the given differential equation. The equation becomes \(1 imes 0 = \frac{1}{2}(0)^2\), simplifying to \(0 = 0\). This confirms \(y_1 = 1\) is a solution.

Step 3: Substitute into the Differential Equation for \(y_2\)

Substitute \(y_2, y_2', y_2''\) into the differential equation. The equation becomes \(x^2 imes 2 = \frac{1}{2}(2x)^2\). This simplifies to \(2x^2 = 2x^2\), confirming \(y_2 = x^2\) is a solution.

Step 4: Check General Solution \(y = c_1 y_1 + c_2 y_2\)

Consider \(y = c_1 \times 1 + c_2 \times x^2 = c_1 + c_2 x^2\). Differentiate to find \(y' = 2c_2 x\) and \(y'' = 2c_2\).

Step 5: Substitute into Differential Equation for General Solution

Substitute \(y\), \(y'\), and \(y''\) The equation becomes \((c_1 + c_2 x^2)(2c_2) = \frac{1}{2}(2c_2 x)^2\). Simplifies to \(2c_2(c_1 + c_2x^2) = 2c_2^2 x^2\). This equality does not hold in general because the terms do not match, indicating \(y = c_1 y_1 + c_2 y_2\) is not a solution.

Key Concepts

Solutions of Differential Equations

A differential equation involves an unknown function and its derivatives. For the given differential equation, the goal is to find functions, known as solutions, that satisfy the equation when substituted for the unknown function. This means that when you plug the solution into the differential equation, it holds true. In our exercise, we had two potential functions, or solutions, to verify:

• \(y_1 = 1\)
• \(y_2 = x^2\)

To verify if these are indeed solutions, we need to check that they satisfy the differential equation. This involves substituting them into the equation along with their derivatives and confirming that both sides become equal.

General Solutions

The concept of a general solution in differential equations refers to a solution that includes arbitrary constants. These constants can be determined if initial conditions or boundary values are provided. However, the exercise revealed something intriguing with regard to forming a general solution. We attempted to create a general solution of the form:

• \(y = c_1 y_1 + c_2 y_2\)
• Where \(c_1\) and \(c_2\) are arbitrary constants.

Despite the individual solutions \(y_1\) and \(y_2\) satisfying the differential equation, their linear combination did not. Typically, in linear differential equations, such a combination would also be a solution, but this is not the case here. This reinforces the importance of checking the behavior of solutions when dealing with nonlinear differential equations.

Verification of Solutions

Verification is a critical step in solving differential equations. This involves substituting a proposed solution back into the original equation to verify that it satisfies the equation. In our exercise, for each proposed solution, we took the following steps:

• Take derivatives of the solution as required by the equation.
• Substitute these derivatives along with the proposed solution into the original differential equation.
• Simplify both sides and check if they equal each other.

Both individual functions \(y_1 = 1\) and \(y_2 = x^2\) passed this test, confirming them as solutions. However, as discussed, the combination \(y = c_1 y_1 + c_2 y_2\) did not satisfy the equation overall, highlighting the importance of verifying each proposed solution or combination carefully.

Derivatives of Functions

Differentiation is a key mathematical operation used to solve differential equations, as it provides the necessary derivatives of a function. In this exercise, understanding how to differentiate our functions clearly was essential.

• For \(y_1 = 1\), derivatives were straightforward as all derivatives of a constant function are zero: \(y_1' = 0\) and \(y_1'' = 0\).
• For \(y_2 = x^2\), derivatives were found using basic power rule of differentiation: \(y_2' = 2x\) and \(y_2'' = 2\).

These derivatives were then used to substitute in the differential equation. Understanding how to correctly differentiate functions ensures clarity and accuracy when verifying solutions, especially when handling more complex or nonlinear differential equations.