Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

Example 21.

1. \(\ds y'' - 2y' + y = e^x \text{.}\)

   Solution.

   First, determine \(y_h\) by solving the characteristic equation:

   \begin{equation*} r^2 - 2r + 1 = 0\text{,} \end{equation*}

   which yields a repeated root \(r_1 = 1\text{,}\) giving the homogeneous solution:

   \begin{equation*} y_h = c_1 e^x + c_2 x e^x\text{.} \end{equation*}

   Since the nonhomogeneous term is \(e^x\text{,}\) the initial guess for the particular solution is:

   \begin{equation*} y_p = A e^x\text{.} \end{equation*}

   However, this form overlaps with \(c_1 e^x\) in \(y_h\text{.}\) To resolve this, multiply \(y_p\) by \(x\text{:}\)

   \begin{equation*} y_p = A x e^x\text{,} \end{equation*}

   which now overlaps with \(c_2 x e^x\text{.}\) Multiplying by \(x\) again gives:

   \begin{equation*} y_p = A x^2 e^x\text{,} \end{equation*}

   which has no terms in common with \(y_h\text{.}\) This is the correct form to use for \(y_p\text{.}\)
2. \(\ds y''' + 4y' = \cos(2x) - \sin(x) \text{.}\)

   Solution.

   First, solve the characteristic equation:

   \begin{equation*} r^3 + 4r = 0\text{,} \end{equation*}

   which yields the roots \(r_1 = 0\text{,}\) \(r_2 = 2i\text{,}\) and \(r_3 = -2i\text{.}\) This results in the homogeneous solution:

   \begin{equation*} y_h = c_1 + c_2 \cos(2x) + c_3 \sin(2x)\text{.} \end{equation*}

   For the nonhomogeneous term \(\cos(2x) - \sin(x)\text{,}\) the initial guess for \(y_p\) is:

   \begin{equation*} y_p = \ub{A \cos(2x) + B \sin(2x)}_{\text{common terms with } y_h} + \ub{D \cos(x) + E \sin(x)}_{\text{NO common terms with } y_h}\text{.} \end{equation*}

   Since \(A \cos(2x)\) and \(B \sin(2x)\) overlap with terms in \(y_h\text{,}\) multiply these terms by \(x\text{:}\)

   \begin{equation*} y_p = \ub{Ax \cos(2x) + B x \sin(2x)}_{\text{now independent}} + \ub{D \cos(x) + E \sin(x)}_{\text{already independent}}\text{.} \end{equation*}

   The terms in \(y_h\) and \(y_p\) are now independent, making this the correct form for \(y_p\text{.}\)
3. \(\ds y'' + y = x e^x \text{.}\)

   Solution.

   Begin by solving the characteristic equation:

   \begin{equation*} r^2 + 1 = 0\text{,} \end{equation*}

   which yields the roots \(r_1 = i\) and \(r_2 = -i\text{,}\) leading to the homogeneous solution:

   \begin{equation*} y_h = c_1 \cos(x) + c_2 \sin(x)\text{.} \end{equation*}

   Since the nonhomogeneous term is \(x e^x\text{,}\) the initial guess for \(y_p\) is:

   \begin{equation*} y_p = (Ax + B) e^x \text{.} \end{equation*}

   This form is not present in \(y_h\text{,}\) so no further adjustment is needed. To determine the coefficients \(A\) and \(B\text{,}\) substitute \(y_p\) back into the differential equation and solve for these values.
4. \(\ds y'' - 6y' + 9y = x^2 e^{3x} \text{.}\)

   Solution.

   First, determine \(y_h\) by solving the characteristic equation:

   \begin{equation*} r^2 - 6r + 9 = 0\text{,} \end{equation*}

   which yields a repeated root \(r = 3\text{,}\) giving the homogeneous solution:

   \begin{equation*} y_h = (c_1 + c_2 x) e^{3x}\text{.} \end{equation*}

   Since the nonhomogeneous term is \(x^2 e^{3x}\text{,}\) the initial guess for the particular solution is:

   \begin{equation*} y_p = (Ax^2 + Bx + C) e^{3x}\text{.} \end{equation*}

   However, notice that \(e^{3x}\) and \(x e^{3x}\) are already present in \(y_h\text{.}\) To avoid overlap, multiply \(y_p\) by \(x^2\text{,}\) giving:

   \begin{equation*} y_p = (Ax^4 + Bx^3 + Cx^2) e^{3x}\text{.} \end{equation*}

   This form is now independent of \(y_h\text{,}\) ensuring that there are no overlapping terms.

   To determine the coefficients \(A\text{,}\) \(B\text{,}\) and \(C\text{,}\) substitute \(y_p\) back into the original differential equation and solve for these values.