Problem 2 - Advanced Engineering Mathematics

Question

In Problems, 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. $$ y y^{\prime \prime}=\frac{1}{2}\left(y^{\prime}\right)^{2} ; y_{1}=1, y_{2}=x^{2} $$

Step-by-Step Solution

Answer

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

Step 1: Verify that \( y_1 = 1 \) is a solution

First, we need to compute the first and second derivatives of \( y_1 \). Since \( y_1 = 1 \) is a constant function, we have \( y_1' = 0 \) and \( y_1'' = 0 \). Substitute these into the differential equation:\[ y_1 y_1'' = \frac{1}{2} (y_1')^2 \]This becomes:\[ 1 \cdot 0 = \frac{1}{2} \cdot 0^2 \]\[ 0 = 0 \]Thus, \( y_1 = 1 \) satisfies the differential equation.

Step 2: Verify that \( y_2 = x^2 \) is a solution

Next, compute the first and second derivatives of \( y_2 \). We have \( y_2' = 2x \) and \( y_2'' = 2 \). Substitute these into the differential equation:\[ y_2 y_2'' = \frac{1}{2} (y_2')^2 \]This becomes:\[ x^2 \cdot 2 = \frac{1}{2} \cdot (2x)^2 \]\[ 2x^2 = \frac{1}{2} \cdot 4x^2 \]\[ 2x^2 = 2x^2 \]Thus, \( y_2 = x^2 \) satisfies the differential equation.

Step 3: Show \( y = c_1 y_1 + c_2 y_2 \) is not a solution

Consider the general solution \( y = c_1 y_1 + c_2 y_2 = c_1 \cdot 1 + c_2 x^2 \). Computing the first and second derivatives:\[ y' = 2c_2 x \]\[ y'' = 2c_2 \]Substitute these into the equation:\[ y y'' = \frac{1}{2} (y')^2 \]\[ (c_1 + c_2 x^2)(2c_2) = \frac{1}{2} (2c_2 x)^2 \]\[ 2c_2(c_1 + c_2 x^2) = 2c_2^2 x^2 \]For general \( c_1, c_2 \), this does not simplify to zero unless specific values are chosen for \( c_1 \) and \( c_2 \), thus \( y \) is not, in general, a solution.

Key Concepts

Solution Verification

When handling differential equations, verifying potential solutions is crucial to ensure they fit the equation's structure. In our exercise, we verified that the functions \( y_1 = 1 \) and \( y_2 = x^2 \) solve the differential equation \( yy'' = \frac{1}{2}(y')^2 \).

To verify solutions, we compute the necessary derivatives and substitute them back into the differential equation. The goal is to check if the equation holds true.

For \( y_1 \), the derivatives are zero, leading to an equation that simply confirms \( 0 = 0 \). Similarly, for \( y_2 \), after substituting the derivatives \( y_2' = 2x \) and \( y_2'' = 2 \), the equation simplifies to \( 2x^2 = 2x^2 \), confirming \( y_2 \) as a solution.

Superposition Principle

The Superposition Principle is a powerful concept, especially when dealing with linear differential equations. It states that if two functions are solutions of a linear differential equation, then any linear combination of these solutions will also be a solution.

However, our exercise demonstrates a case where the differential equation is not linear, specifically because of the multiplying terms on \( y \) and its derivatives. In this context, the general form \( y = c_1 y_1 + c_2 y_2 \) (where \( c_1 \) and \( c_2 \) are constants) is generally not a solution to the equation. This deviation occurs because the differential equation includes nonlinear terms, violating the conditions needed for the superposition principle to hold.

Mathematical Derivatives

Derivatives are fundamental tools in calculus, showing how a function changes as its inputs change. They are prominently used in differential equations to describe the dynamics of systems.

In our exercise, we computed first and second derivatives to verify if \( y_1 = 1 \) and \( y_2 = x^2 \) are solutions. For \( y_1 \), being a constant, both derivatives are zero. Derivative computations for \( y_2 \) yielded \( y_2' = 2x \) and \( y_2'' = 2 \).

The accurate calculation of derivatives is essential for substitution back into the differential equation, which helps verify potential solutions. Remember, derivatives focus on the rate of change, offering precious insights into the behavior of functions.

Non-Linearity of Solutions

Non-linearity in differential equations introduces complexities that can prevent the application of linear solution techniques such as the Superposition Principle. Our given differential equation involves products of terms, like \( yy'' \), that make it nonlinear.

When we tried to show the general solution \( y = c_1 y_1 + c_2 y_2 \) was not a solution, the non-linear nature of the problem became evident. Even though each function individually satisfied the equation, their linear combination did not.

Understanding non-linearity is crucial when solving or verifying solutions to differential equations, as it significantly affects the solution space and methods applicable to find or verify solutions.