Problem 35 - Advanced Engineering Mathematics

Question

(a) Verify that \(y_{p_{1}}=3 e^{2 x}\) and \(y_{p_{2}}=x^{2}+3 x\) are, respectively, particular solutions of $$ y^{\prime \prime}-6 y^{\prime}+5 y=-9 e^{2 x} $$ and \(\quad y^{\prime \prime}-6 y^{\prime}+5 y=5 x^{2}+3 x-16\). (b) Use part (a) to find particular solutions of $$ y^{\prime \prime}-6 y^{\prime}+5 y=5 x^{2}+3 x-16-9 e^{2 x} $$ and \(y^{\prime \prime}-6 y^{\prime}+5 y=-10 x^{2}-6 x+32+e^{2 x}\).

Step-by-Step Solution

Answer

Solve using combination: \( y_p = 3e^{2x} + x^2 + 3x \) and \( y_p = -3e^{2x} - x^2 - 3x \).

Step 1: Verify particular solution for first equation

To verify that \(y_{p_{1}}=3 e^{2x}\) is a particular solution to \(y'' - 6y' + 5y = -9e^{2x}\), start by finding \(y'_{p_{1}}\) and \(y''_{p_{1}}\): \[ y'_{p_{1}} = \frac{d}{dx}(3e^{2x}) = 6e^{2x} \]\[ y''_{p_{1}} = \frac{d}{dx}(6e^{2x}) = 12e^{2x} \]Now substitute \(y_{p_{1}}, y'_{p_{1}},\) and \(y''_{p_{1}}\) into the differential equation:\[ 12e^{2x} - 6(6e^{2x}) + 5(3e^{2x}) = -9e^{2x} \]\[ 12e^{2x} - 36e^{2x} + 15e^{2x} = -9e^{2x} \]\[ -9e^{2x} = -9e^{2x} \]The identity holds, so \(y_{p_{1}}=3e^{2x}\) is indeed a particular solution.

Step 2: Verify particular solution for second equation

Next, verify \(y_{p_{2}}=x^{2}+3x\) is a particular solution for \(y'' - 6y' + 5y = 5x^{2} + 3x - 16\). Calculate \(y'_{p_{2}}\) and \(y''_{p_{2}}\):\[ y'_{p_{2}} = \frac{d}{dx}(x^2 + 3x) = 2x + 3 \]\[ y''_{p_{2}} = \frac{d}{dx}(2x + 3) = 2 \]Substitute \(y_{p_{2}}, y'_{p_{2}},\) and \(y''_{p_{2}}\) into the differential equation:\[ 2 - 6(2x + 3) + 5(x^2 + 3x) = 5x^2 + 3x - 16 \]\[ 2 - 12x - 18 + 5x^2 + 15x = 5x^2 + 3x - 16 \]\[ 5x^2 + 3x - 16 = 5x^2 + 3x - 16 \]The equation holds, confirming \(y_{p_{2}}=x^{2}+3x\) is a particular solution.

Step 3: Solve for combined particular solution

To find a particular solution for \(y'' - 6y' + 5y = 5x^{2} + 3x - 16 - 9e^{2x}\), combine \(y_{p_{1}}\) and \(y_{p_{2}}\):\[ y_p = y_{p_{1}} + y_{p_{2}} = 3e^{2x} + (x^{2} + 3x) \]Thus, \(y_p = 3e^{2x} + x^{2} + 3x\) is a particular solution for the given equation.

Step 4: Verify particular solution for new equation

From part (a), to find a particular solution for \(y'' - 6y' + 5y = -10x^{2} - 6x + 32 + e^{2x}\), initially note the existing solutions:- Correct sign of \(y_{p_{1}}\) as particular solution.- Adjust \(y_{p_{2}}\) per new coefficients.Thus, for combined actions: Combine the opposite of \(y_{p_{2}} = x^2 + 3x\) with the adjusted sign of \(y_{p_{1}} = -3e^{2x}\) derived individually to form:\[ y_p = -3e^{2x} - x^{2} - 3x \]

Key Concepts

Particular Solutions

In differential equations, a particular solution represents a specific solution to a differential equation that fits the entirety of a given non-homogeneous equation. It's a form of a solution that satisfies the full conditions of the equation, including any additional terms beyond the homogeneous part. For example, the solutions in our exercise, \(y_{p_{1}}\) and \(y_{p_{2}}\), are particular because they not only satisfy the differential equation's structure but also account for unique functions like \(-9e^{2x}\) and \(5x^2 + 3x - 16\). These are distinct from general solutions, which represent a family of solutions for the corresponding homogeneous equation without considering such additional terms.

Second Order Differential Equations

Second-order differential equations, such as \(y'' - 6y' + 5y = 0\), require both first and second derivatives of a function. They are fundamental to modeling real-world phenomena involving acceleration or curvature, such as in physics and engineering. In these problems, we often deal with solutions where we must verify particular functions that satisfy the equations' non-homogeneous parts.

Second-order linear differential equations can have both homogeneous and non-homogeneous components. The homogeneous part is characterized by the absence of any external influences, while the non-homogeneous one includes these additional terms. Solving these involves finding both the complementary function, derived from the homogeneous part, and particular solutions that satisfy the full equation.

Exponential Functions

Exponential functions are a crucial part of solutions to differential equations, particularly when involved terms have exponential behavior. In the exercise, \(3e^{2x}\) is used as a particular solution. The exponential function \(e^{2x}\) increases rapidly, making it a powerful form when modeling growth processes or decay.In the context of differential equations:

• Exponential functions often appear in solutions due to their properties when differentiated.
• They simplify calculations because the differentiation of \(e^{ax}\) results in itself scaled by the constant \(a\).

This unique characteristic allows exponential functions to work seamlessly as solutions or components of solutions in linear differential equations.

Polynomial Solutions

Polynomial solutions involve polynomial expressions which can satisfy differential equations that have polynomial terms. In the solved exercise, \(x^2 + 3x\) serves as a particular solution, specifically designed to match the polynomial terms on the non-homogeneous side of the equation, \(5x^2 + 3x - 16\).Polynomials are beneficial in such problems due to:

• Their straightforward differentiation process, which reduces step by step.
• Matching polynomial forms can handle non-homogeneous terms, making them excellent candidates for particular solutions.

Overall, polynomial solutions are integral in cases where the differential equation presents polynomial characteristics, and they help seamlessly integrate these terms into the broader solution framework.