Problem 43 - Advanced Engineering Mathematics

Question

Consider the differental equation \(a y^{\prime \prime}+b y^{\prime}+c y=e^{k x}\), where \(a, b, c\), and \(k\) are constants. The auxiliary equation of the associated homogeneous equation is $$ a m^{2}+b m+c=0 $$ (a) If \(k\) is not a root of the auxiliary equation, show that we can find a particular solution of the form \(y_{p}=A e^{k x}\), where \(A=1 /\left(a k^{2}+b k+c\right)\) (b) If \(k\) is a root of the auxiliary equation of multiplicity one, show that we can find a particular solution of the form \(y_{p}=A x e^{k x}\), where \(A=1 /(2 a k+b) .\) Explain how we lenow that \(k \neq-b /(2 a)\) (c) If \(k\) is a root of the auxiliary equation of multiplicity two, show that we can find a particular solution of the form \(y=A x^{2} e^{k x}\), where \(A=1 /(2 a)\)

Step-by-Step Solution

Answer

(a) A particular solution is \( y_p = \frac{e^{kx}}{ak^2 + bk + c} \). (b) Use \( y_p = \frac{x e^{kx}}{2ak + b} \) assuming \( k \neq -\frac{b}{2a} \). (c) Solution is \( y_p = \frac{x^2 e^{kx}}{2a} \).

Step 1: Understanding Part (a)

For part (a), consider the differential equation \[ a y^{\prime \prime} + b y^{\prime} + c y = e^{kx} \]and its associated homogeneous equation:\[ a y^{\prime \prime} + b y^{\prime} + c y = 0 \]The solution for the homogeneous equation is based on the auxiliary equation \( a m^2 + b m + c = 0 \). We need a particular solution since \( k \) is not a root of the auxiliary equation, which implies that \( a k^2 + b k + c eq 0 \).

Step 2: Finding Particular Solution for Part (a)

Assume a particular solution of the form \( y_p = A e^{k x} \) and substitute into the differential equation:\[ a(Ak^2 e^{kx}) + b(Ak e^{kx}) + c(A e^{kx}) = e^{kx} \]Factor out \( A e^{kx} \):\[ A (ak^2 + bk + c) e^{kx} = e^{kx} \]Equate coefficients and solve for \( A \):\[ A (ak^2 + bk + c) = 1 \]Thus, \( A = \frac{1}{ak^2 + bk + c} \).

Step 3: Understanding Part (b)

For part (b), if \( k \) is a root with multiplicity one, the corresponding homogeneous solution involves \( e^{kx} \). This implies:\[ a k^2 + b k + c = 0 \]The particular solution should thus involve \( x e^{kx} \) to avoid overlap with the homogeneous solution, so assume \( y_p = A x e^{kx} \).

Step 4: Finding Particular Solution for Part (b)

Substituting \( y_p = A x e^{kx} \) into the DE:\[ (A'x + A)ak^2 e^{kx} + (A' + Akx)bk e^{kx} + Acx e^{kx} = e^{kx} \]Simplifying equation:\[ Ax(ak^2 + bk + c)e^{kx} + A(-2ak - b)e^{kx} = e^{kx} \]Equate coefficients to find \( A \):\[ A = \frac{1}{2ak + b} \]

Step 5: Verifying Condition for Part (b)

Ensure \( 2ak + b eq 0 \) since \( e^{kx} \) is part of the homogeneous solution. \( k \) cannot satisfy \( 2ak + b = 0 \), as it would imply an ill-posed solution set with a division by zero.

Step 6: Understanding Part (c)

For part (c), \( k \) is a double root of the auxiliary equation, then:\[ a m^2 + b m + c = 0 \Rightarrow am^2 + bm + c = (m-k)^2 \]Assume a particular solution of the form \( y_p = A x^2 e^{kx} \) to avoid redundancy and meet higher root multiplicity.

Step 7: Finding Particular Solution for Part (c)

Substitute \( y_p = A x^2 e^{kx} \) into the differential equation:\[ 2Ax ak^2 e^{kx} + A (2xk^2 + 4kx)e^{kx} + 2Axk^2e^{kx} = e^{kx} \]Extract terms involving \( e^{kx} \):\[ Ax^2(ak^2)e^{kx} = e^{kx} \]From this, \( A = \frac{1}{2a} \) is derived by matching coefficients.

Key Concepts

Auxiliary Equation

The auxiliary equation plays a crucial role in solving linear differential equations with constant coefficients. It is associated with the homogeneous version of a differential equation, where all terms on one side equal zero. For an equation of the form \[ a y'' + b y' + c y = e^{kx} \]the corresponding homogeneous equation is\[ a y'' + b y' + c y = 0 \]. The auxiliary equation is obtained by replacing derivatives \(y''\) with \(m^2\), \(y'\) with \(m\), and the function \(y\) with \(1\):\[ a m^2 + b m + c = 0 \]This quadratic equation helps determine the complementary (or homogeneous) solution. The nature of its roots (real/complex, distinct or repeated) influences the form of this solution, which is essential for constructing the general solution of the differential equation.

Particular Solution

A particular solution addresses the non-homogeneous part of a differential equation such as \[ a y'' + b y' + c y = e^{kx} \]. The approach to finding it depends largely on whether \(k\) is a root of the auxiliary equation. The goal is to identify a function, \( y_p \), that satisfies the original differential equation. If \(k\) is not a root of the auxiliary equation, a straightforward particular solution is\[ y_p = A e^{kx} \],where\[ A = \frac{1}{a k^2 + b k + c} \]. This form ensures that \(y_p\) does not overlap with the homogeneous solution, which is based on the distinct roots of the auxiliary equation. If \(k\) is a root of the auxiliary equation, alternative forms, such as \(y_p = Ax e^{kx}\) (for a simple root), or \(y_p = Ax^2 e^{kx}\) (for a double root), are used to avoid redundancy and ensure a valid solution.

Homogeneous Equation

The homogeneous equation is formed when the non-homogeneous part of the original equation is set to zero. For the differential equation \[ a y'' + b y' + c y = e^{kx} \],the associated homogeneous equation is simply \[ a y'' + b y' + c y = 0 \]. Solving this provides the complementary solution, a fundamental component in constructing the general solution of the differential equation.The solution involves roots of the auxiliary equation \[ a m^2 + b m + c = 0 \]. These roots, termed as characteristic roots, guide the form of the homogeneous solution:

• If roots are real and distinct, the solution is \(C_1 e^{m_1 x} + C_2 e^{m_2 x}\).
• If roots are real and repeated, \(C_1 e^{m x} + C_2 x e^{m x}\).
• For complex roots \(\alpha \pm \beta i\), the solution is \(e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x))\).

Understanding this structure is key to solving more complex differential equations effectively.