Harmonic Oscillator Equation Analysis

The given equation is a second-order linear non-homogeneous differential equation with constant coefficients, representing a one-dimensional forced harmonic oscillator with a dissipative force:

\(\frac{{{{\rm{d}}^{\rm{2}}}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^{\rm{2}}}}}\,{\rm{ + }}\,{\rm{10}}\frac{{{\rm{dx}}}}{{{\rm{dt}}}}\,{\rm{ + }}\,{\rm{16x}}\,{\rm{ = }}\,{\rm{6t}}{{\rm{e}}^{{\rm{ - 8t}}}}{\rm{ + }}\,{\rm{4}}{{\rm{t}}^{\rm{2}}}{{\rm{e}}^{{\rm{ - 2t}}}}\)

This equation is in the form \(ay'' + by' + cy = g(t)\), where \(a=1\), \(b=10\), \(c=16\), and the non-homogeneous term (forcing function) is \(g(t) = 6te^{-8t} + 4t^2e^{-2t}\).

To find the general form of the particular solution \(x_p(t)\), we use the method of undetermined coefficients. This method requires us to analyze the non-homogeneous term \(g(t)\) and the roots of the characteristic equation of the corresponding homogeneous equation.

Characteristic Equation and Roots

The corresponding homogeneous equation is \(x'' + 10x' + 16x = 0\). The characteristic equation is obtained by replacing \(x''\) with \(r^2\), \(x'\) with \(r\), and \(x\) with \(1\):

\(r^2 + 10r + 16 = 0\)

We solve this quadratic equation for \(r\):

\((r+2)(r+8) = 0\)

The roots are \(r_1 = -2\) and \(r_2 = -8\). Both roots are real and distinct.

Particular Solution using Undetermined Coefficients

The non-homogeneous term is a sum of two functions: \(g_1(t) = 6te^{-8t}\) and \(g_2(t) = 4t^2e^{-2t}\). We find the particular solution for each term separately and sum them up.

Analyzing the Term \(g_1(t) = 6te^{-8t}\)

This term is of the form \(P_m(t)e^{\alpha t}\), where \(P_m(t) = 6t\) is a polynomial of degree \(m=1\) and \(\alpha = -8\).

We compare \(\alpha = -8\) with the roots of the characteristic equation (\(-2\) and \(-8\)). Since \(\alpha = -8\) is equal to one of the roots (\(r_2 = -8\)), and this root has a multiplicity of 1, the initial guess for the particular solution for this term must be multiplied by \(t^s\), where \(s\) is the multiplicity of the root equal to \(\alpha\).

Here, \(s=1\).

The general form of the polynomial for the guess is a polynomial of the same degree as \(P_m(t)\), which is degree 1. Let this general polynomial be \(Dt + E\).

So, the form of the particular solution component for \(g_1(t)\) is:

\(x_{p1}(t) = t^1 (Dt + E)e^{-8t} = t(Dt + E)e^{-8t}\)

Analyzing the Term \(g_2(t) = 4t^2e^{-2t}\)

This term is of the form \(Q_n(t)e^{\beta t}\), where \(Q_n(t) = 4t^2\) is a polynomial of degree \(n=2\) and \(\beta = -2\).

We compare \(\beta = -2\) with the roots of the characteristic equation (\(-2\) and \(-8\)). Since \(\beta = -2\) is equal to one of the roots (\(r_1 = -2\)), and this root has a multiplicity of 1, the initial guess for the particular solution for this term must be multiplied by \(t^s\), where \(s\) is the multiplicity of the root equal to \(\beta\).

Here, \(s=1\).

The general form of the polynomial for the guess is a polynomial of the same degree as \(Q_n(t)\), which is degree 2. Let this general polynomial be \(At^2 + Bt + C\).

So, the form of the particular solution component for \(g_2(t)\) is:

\(x_{p2}(t) = t^1 (At^2 + Bt + C)e^{-2t} = t(At^2 + Bt + C)e^{-2t}\)

General Form of the Particular Solution

The general form of the particular solution \(x_p(t)\) is the sum of the forms derived for \(g_1(t)\) and \(g_2(t)\):

\(x_p(t) = x_{p1}(t) + x_{p2}(t)\)

\(x_p(t) = t(Dt + E)e^{-8t} + t(At^2 + Bt + C)e^{-2t}\)

Rearranging the terms, we get:

\(x_p(t) = t(At^2 + Bt + C)e^{-2t} + t(Dt + E)e^{-8t}\)

Comparing with Options

Let's compare this derived form with the given options:

• Option 1: t(At² + Bt + C)e^-2t + (Dt + E)e^-8t - Incorrect (second term missing factor of t)
• Option 2: (At² + Bt + C)e^-2t + (Dt + E)e^-8t - Incorrect (both terms missing factor of t)
• Option 3: t(At² + Bt + C)e^-2t + t(Dt + E)e^-8t - Correct (matches the derived form)
• Option 4: (At² + Bt + C)e^-2t + t(Dt + E)e^-8t - Incorrect (first term missing factor of t)

The general form of the particular solution is indeed \(t(At^2 + Bt + C)e^{-2t} + t(Dt + E)e^{-8t}\).