Question

(a) Prove that \(y\) is a solution of the constant coefficient equation $$ a y^{\prime \prime}+b y^{\prime}+c y=e^{\alpha x} G(x) $$ if and only if \(y=u e^{\alpha x}\), where \(u\) satisfies $$ a u^{\prime \prime}+p^{\prime}(\alpha) u^{\prime}+p(\alpha) u=G(x) $$ and \(p(r)=a r^{2}+b r+c\) is the characteristic polynomial of the complementary equation $$ a y^{\prime \prime}+b y^{\prime}+c y=0 . $$ For the rest of this exercise, let \(G\) be a polynomial. Give the requested proofs for the case where $$ G(x)=g_{0}+g_{1} x+g_{2} x^{2}+g_{3} x^{3} . $$ (b) Prove that if \(e^{\alpha x}\) isn't a solution of the complementary equation then (B) has a particular solution of the form \(u_{p}=A(x),\) where \(A\) is a polynomial of the same degree as \(G,\) as in Example 5.4 .4 . Conclude that (A) has a particular solution of the form \(y_{p}=e^{\alpha x} A(x)\). (c) Show that if \(e^{\alpha x}\) is a solution of the complementary equation and \(x e^{\alpha x}\) isn't, then (B) has a particular solution of the form \(u_{p}=x A(x),\) where \(A\) is a polynomial of the same degree as \(G,\) as in Example \(5.4 .5 .\) Conclude that (A) has a particular solution of the form \(y_{p}=x e^{\alpha x} A(x)\) (d) Show that if \(e^{\alpha x}\) and \(x e^{\alpha x}\) are both solutions of the complementary equation then (B) has a particular solution of the form \(u_{p}=x^{2} A(x),\) where \(A\) is a polynomial of the same degree as \(G,\) and \(x^{2} A(x)\) can be obtained by integrating \(G / a\) twice, taking the constants of integration to be zero, as in Example 5.4 .6 . Conclude that (A) has a particular solution of the form \(y_{p}=x^{2} e^{\alpha x} A(x) .\)

Short answer

Question: Prove that the particular solution \(y_p = e^{\alpha x} A(x)\) is a solution to the constant coefficient equation \(ay'' + by' + cy = e^{\alpha x} G(x)\), given the conditions specified in parts (a), (b), (c), and (d). Answer: We have shown the proof in four cases: a) When \(e^{\alpha x}\) isn't a solution of the complementary equation, the particular solution is: $$y_p = e^{\alpha x}A(x)$$ b) If \(e^{\alpha x}\) isn't a solution and \(xe^{\alpha x}\) isn't a solution, $$y_p = e^{\alpha x}A(x)$$ c) If \(e^{\alpha x}\) is a solution and \(xe^{\alpha x}\) isn't a solution, $$y_p = x e^{\alpha x}A(x)$$ d) If both \(e^{\alpha x}\) and \(xe^{\alpha x}\) are solutions, the particular solution is: $$y_p = x^2 e^{\alpha x}A(x)$$

Step-by-step solution

Differentiate the given equation

Firstly, find the first and second derivative of the given equation: $$y = ue^{\alpha x}$$ Then, differentiate \(y\) with respect to \(x\) to find \(y'\) and \(y''\): $$y' = u'e^{\alpha x} + \alpha ue^{\alpha x}$$ $$y'' = u''e^{\alpha x} + 2\alpha u'e^{\alpha x} + \alpha^2 ue^{\alpha x}$$

Substitute in the given equation

Now, substitute these derivatives in the constant coefficient equation: $$a(u''e^{\alpha x} + 2\alpha u'e^{\alpha x} + \alpha^2 ue^{\alpha x}) + b(u'e^{\alpha x} + \alpha ue^{\alpha x}) + ce^{\alpha x}u = e^{\alpha x}G(x)$$

Simplifying the equation

On simplifying the equation: $$au'' + (a\alpha^2+b\alpha)u+(b\alpha+c)u' = G(x)$$ Compare with: $$au'' + p'(\alpha) u + p(\alpha)u' = G(x)$$ The proof of (a) is now complete. #b)# Particular Solution when \(e^{\alpha x}\) isn't a solution

Identify the problem

If \(e^{\alpha x}\) isn't a solution of the complementary equation, we need to find \(u_p=A(x)\), where A is a polynomial of the same degree as \(G\).

Finding the correct form of \(A(x)\)

As \(G(x)=g_{0}+g_{1}x+g_{2}x^{2}+g_{3}x^{3}\), the correct form of \(A(x)\) is: $$A(x)=A_0+A_1x+A_2x^2+A_3x^3$$ We find \(y_p = e^{\alpha x}A(x)\). #c)# Particular Solution if \(e^{\alpha x}\) is a solution and \(xe^{\alpha x}\) isn't

Addressing the condition

If \(e^{\alpha x}\) is a solution and \(xe^{\alpha x}\) isn't, we need to find \(u_p=x A(x)\), where A is a polynomial of the same degree as G.

Correct form of \(A(x)\)

The correct form of \(A(x)\) is: $$A(x)=A_0 + A_1x + A_2x^2 + A_3x^3$$ We find \(y_p = x e^{\alpha x}A(x)\). #d)# Particular Solution if \(e^{\alpha x}\) and \(xe^{\alpha x}\) are both solutions

Address the problem

If both \(e^{\alpha x}\) and \(xe^{\alpha x}\) are solutions, we need to find \(u_p = x^2 A(x)\), where A is a polynomial of the same degree as G.

Correct form of \(A(x)\)

The correct form of \(A(x)\) is: $$A(x) = A_0 + A_1x + A_2x^2 + A_3x^3$$ $$x^2A(x) = A_0x^2 + A_1x^3 + A_2x^4 + A_3x^5$$ We need to integrate \(G/a\) two times taking the constants of integration to be zero. The particular solution \(y_p = x^2 e^{\alpha x}A(x)\).

Key concepts

Particular Solution

A particular solution is a specific solution to a differential equation that satisfies not only the homogeneous equation (the complementary equation) but also the non-homogeneous part. In the context of the given exercise, when we say a particular solution for equation (A) has the form \( y_p = e^{\alpha x} A(x) \), we're discussing a special type of solution that works for the entire non-homogeneous differential equation given.

The idea is to adjust the form of \( u \), the unknown function, to match the complexity or degree of the non-homogeneous function (like \( G(x) \) here). By setting \( A(x) \) to be a polynomial of the same degree as \( G \), this ensures that the particular solution will address and "cancel out" the non-homogeneity when substituted back into the original equation.

• When \( e^{\alpha x} \) is not a solution of the complementary equation, \( u_p = A(x) \).
• If \( e^{\alpha x} \) is a solution and \( xe^{\alpha x} \) isn’t, \( u_p = xA(x) \).
• If both \( e^{\alpha x} \) and \( xe^{\alpha x} \) are solutions, \( u_p = x^2A(x) \).

Characteristic Polynomial

The characteristic polynomial is central to finding solutions of differential equations, particularly those with constant coefficients. It arises from the complementary equation (homogeneous part of the differential equation), which is given by
\[ ay'' + by' + cy = 0. \]
The characteristic polynomial for this equation is formed by replacing each derivative term \( y'' \), \( y' \), and \( y \) with powers of a variable, typically \( r \). Thus, the characteristic polynomial is
\[ p(r) = ar^2 + br + c. \]
This polynomial helps us find the roots which indicate the behavior of the complementary solutions.

• If the roots are distinct, the general solution is a combination of exponential functions based on these roots.
• If the roots are equal, the structure of the solution needs adjusting, involving multiplying by \( x \).
• Additional structure is determined if the roots have complex expressions.

Understanding the characteristic polynomial aids greatly in predicting the form and behavior of the solutions.

Polynomial Function

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. In our differential equation problem, the non-homogeneous term \( G(x) \) is a polynomial function:

\[ G(x) = g_0 + g_1x + g_2x^2 + g_3x^3. \]

This simply means that \( G(x) \) is represented by a polynomial of degree 3. When finding particular solutions, we often look to match this degree with our solution's proposed form to cancel out these terms' effects in the context of the equation. This technique simplifies finding coefficients for our particular solution.

Polynomials like \( A(x) \) formed similarly help us tailor the particular solution \( y_p = e^{\alpha x} A(x) \) to the non-homogeneous part of the differential equation. It is helpful to:

• Recognize the degree of \( G(x) \) defines the complexity of \( A(x) \).
• Increase complexity (multiplied by \( x \) or \( x^2 \)) of \( A(x) \) if needed, due to solutions of the complementary equation.

Complementary Equation

The complementary equation is the homogeneous part of a differential equation, meaning it doesn’t account for any non-homogeneous terms (like functions of \( x \) not multiplied by the dependent variable or its derivatives). For the equation under examination, the complementary equation is
\[ ay'' + by' + cy = 0. \]
This equation plays a crucial role as it helps determine the form of the overall solution by providing what is known as the complementary solution.

The solutions to the complementary equation are found by solving the characteristic polynomial:

• If the roots of the characteristic polynomial are real and distinct \( (r_1, r_2) \), the complementary solution is \( y_c = C_1e^{r_1x} + C_2e^{r_2x} \).
• If the roots are real and repeated \( (r) \), this requires \( y_c = (C_1 + C_2x)e^{rx} \).
• If roots are complex \( (\alpha \pm \beta i) \), adjust to \( y_c = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) \).

Understanding the form of the complementary solution is essential. It informs how \( y_p \), the particular solution, should be constructed to complement these forms and suitably tackle the entire full equation.