Example D.9.

Example D.10.

Example D.11.

Example D.12.

Example D.13.

1. \begin{equation*} g\left( t \right) = 16{{\bf{e}}^{7t}}\sin \left( {10t} \right) \end{equation*}
2. \begin{equation*} g\left( t \right) = \left( {9{t^2} - 103t} \right)\cos t \end{equation*}
3. \begin{equation*} g\left( t \right) = - {{\bf{e}}^{ - 2t}}\left( {3 - 5t} \right)\cos \left( {9t} \right) \end{equation*}

Example D.14.

1. \begin{equation*} g\left( t \right) = 4\cos \left( {6t} \right) - 9\sin \left( {6t} \right) \end{equation*}
2. \begin{equation*} g\left( t \right) = - 2\sin t + \sin \left( {14t} \right) - 5\cos \left( {14t} \right) \end{equation*}
3. \begin{equation*} g\left( t \right) = {{\bf{e}}^{7t}} + 6 \end{equation*}
4. \begin{equation*} g\left( t \right) = 6{t^2} - 7\sin \left( {3t} \right) + 9 \end{equation*}
5. \begin{equation*} g\left( t \right) = 10{{\bf{e}}^t} - 5t{{\bf{e}}^{ - 8t}} + 2{{\bf{e}}^{ - 8t}} \end{equation*}
6. \begin{equation*} g\left( t \right) = {t^2}\cos t - 5t\sin t \end{equation*}
7. \begin{equation*} g\left( t \right) = 5{{\bf{e}}^{ - 3t}} + {{\bf{e}}^{ - 3t}}\cos \left( {6t} \right) - \sin \left( {6t} \right) \end{equation*}

Example D.15.

Example D.16.

1. \(\ds y'' + 3y' + 2y = 6e^{2x} \text{.}\)

   Solution.

   To select the appropriate form for \(y_p\text{,}\) we should find \(y_h\) to ensure that \(y_p\) and \(y_h\) have no terms in common. Solving the LHCC version of our equation, we find

   \begin{equation*} y_h = c_1 e^{-x} + c_2 e^{-2x}\text{.} \end{equation*}

   Next, we select the initial form of \(y_p\text{.}\) Since the right-hand side is an \(e^{2x}\) function, we initially set \(y_p = A e^{2x}\) and note that \(y_h\) has no \(e^{2x}\) terms, so we proceed to find \(A\) by calculating two derivatives of \(y_p\text{,}\)

   \begin{align*} y_p =\amp\ A e^{2x} \\ y_p' =\amp\ 2A e^{2x} \\ y_p'' =\amp\ 4A e^{2x} \text{,} \end{align*}

   and substituting these into the equation and collecting like terms, to get

   \begin{gather*} 4A e^{2x} + 3(2A e^{2x}) + 2(A e^{2x}) = 6e^{2x} \\ 4A e^{2x} + 6A e^{2x} + 2A e^{2x} = 6e^{2x} \\ \ul{12A} e^{2x} = \ul{6} e^{2x} \text{.} \end{gather*}

   Matching the underlined coefficients leads to the equation

   \begin{gather*} 12A = 6 \quad \implies \quad A = \frac{1}{2} \text{.} \end{gather*}

   Finally, we write the particular solution as

   \begin{equation*} y_p = \frac{1}{2} e^{2x} \text{.} \end{equation*}
2. \(\ds y'' - 3y' + y = 2x^2 + 3x \text{.}\)

   Solution.

   As before, we first find the homogeneous solution, which is

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

   Next, we select the initial form of \(y_p\) as

   \begin{gather*} y_p = Ax^2 + Bx + C \end{gather*}

   and note that \(y_p\) has no terms in common with \(y_h\text{,}\) so we compute \(y_p'\text{,}\) \(y_p''\text{,}\) plug these into the equation, and collect the \(x^2\text{,}\) \(x\text{,}\) and constant terms to get

   \begin{equation*} y_p' = 2Ax + B \qquad y_p'' = 2A \end{equation*}

   \begin{gather*} 2A - 3(2Ax + B) + (Ax^2 + Bx + C) = 2x^2 + 3x \\ 2A - 6Ax - 3B + Ax^2 + Bx + C = 2x^2 + 3x \\ \us{1}{\ul{A}}x^2 + \us{2}{\ul{(- 6A + B)}}x + \us{3}{\ul{(2A - 3B + C)}} = \us{1}{\ul{2}}x^2 + \us{2}{\ul{3}}x + \us{3}{\ul{0}} \text{.} \end{gather*}

   Matching the underlined coefficients leads to the following system of equations and solution for the coefficients \(A\text{,}\) \(B\text{,}\) and \(C\text{:}\)

   \begin{align*} \amp\ 1. \\ \amp\ 2. \\ \amp\ 3. \end{align*}

   \begin{align*} A =\amp\ 2 \\ -6A + B =\amp\ 3 \\ 2A - 3B + C =\amp\ 0 \end{align*}

   \begin{gather*} \\ \rightarrow \\ \end{gather*}

   \begin{align*} \amp\ A = 2 \\ \amp\ B = 9 \\ \amp\ C = 13 \end{align*}

   Finally, we write the particular solution as

   \begin{equation*} y_p = 2x^2 + 9x + 13 \text{.} \end{equation*}
3. \(\ds y'' - 4y' - 12y = t{e^{4t}} \text{.}\)

   Solution.

   First find the homogeneous solution,

   \begin{equation*} y_h = c_1 e^{6t} + c_2 e^{-2t} \text{.} \end{equation*}

   For this example we need to select a particular solution that is a polynomial times an exponential. So, we’ll guess that the particular solution is

   \begin{equation*} Y_P\left( t \right) = \left( {At + B} \right){e^{4t}} \end{equation*}

   Now, we need to compute the first and second derivatives of this and plug them into the differential equation. Doing this gives

   \begin{gather*} Y_P'\left( t \right) = A{e^{4t}} + 4At{e^{4t}} + B{e^{4t}}\\ Y_P''\left( t \right) = 4A{e^{4t}} + 4A{e^{4t}} + 16At{e^{4t}} + 4B{e^{4t}} \end{gather*}

   Substituting these into the differential equation gives

   \begin{align*} 4A{e^{4t}} \amp\ + 4A{e^{4t}} + 16At{e^{4t}} + 4B{e^{4t}}\\ \amp\ - 4\left( {A{e^{4t}} + 4At{e^{4t}} + B{e^{4t}}} \right) - 12\left( {At + B} \right){e^{4t}} = t{e^{4t}} \end{align*}

   Grouping the \(e^{4t}\) and \(te^{4t}\) terms together gives us

   \begin{gather*} 16At{e^{4t}} - 4\left( {4At{e^{4t}}} \right) - 12At{e^{4t}} = t{e^{4t}}\\ 0 = t{e^{4t}} \end{gather*}

   Matching coefficients gives us \(A = \frac{1}{4}\text{.}\) So, the particular solution is

   \begin{equation*} Y_P\left( t \right) = \left( {\frac{1}{4}t + B} \right){e^{4t}} \end{equation*}

   We now need to find \(B\text{.}\) To do this we’ll need to plug this into the differential equation and solve for \(B\text{.}\) Doing this gives

   \begin{gather*} 4\left( {\frac{1}{4}t + B} \right){e^{4t}} + 4\left( {\frac{1}{4}t + B} \right){e^{4t}} + 16\left( {\frac{1}{4}t + B} \right)t{e^{4t}} + 4B{e^{4t}} - 4\left( {\frac{1}{4}{e^{4t}} + 4\left( {\frac{1}{4}t + B} \right){e^{4t}} + B{e^{4t}}} \right) - 12\left( {\frac{1}{4}t + B} \right){e^{4t}} = t{e^{4t}}\\ {e^{4t}} + {e^{4t}} + 4t{e^{4t}} + 4B{e^{4t}} - {e^{4t}} - 4t{e^{4t}} - 4B{e^{4t}} - 3t{e^{4t}} - 3B{e^{4t}} = t{e^{4t}}\\ - 3B{e^{4t}} = 0 \end{gather*}

   Matching coefficients gives us \(B = 0\text{.}\) So, the particular solution is then

   \begin{equation*} Y_P\left( t \right) = \frac{1}{4}t{e^{4t}} \end{equation*}

   A particular solution for this differential equation is then

   \begin{equation*} {Y_P}\left( t \right) = {e^{4t}}\left( { - \frac{t}{{12}} - \frac{1}{{36}}} \right) = - \frac{1}{{36}}\left( {3t + 1} \right){e^{4t}} \end{equation*}
4. \(\ds y'' - 4y' - 12y = 3{e^{5t}} + \sin \left( {2t} \right) + t{e^{4t}} \text{.}\)

   Solution.

   This example is the reason that we’ve been using the same homogeneous differential equation for all the previous examples. There is nothing to do with this problem. All that we need to do it go back to the appropriate examples above and get the particular solution from that example and add them all together.

   Doing this gives

   \begin{equation*} {Y_P}\left( t \right) = - \frac{3}{7}{e^{5t}} + \frac{1}{{40}}\cos \left( {2t} \right) - \frac{1}{{20}}\sin \left( {2t} \right) - \frac{1}{{36}}\left( {3t + 1} \right){e^{4t}} \end{equation*}
5. \(\ds y'' + y = 3 \cos x \text{.}\)

   Solution.

   Step 1: The right-hand side is trigonometric, so we guess:

   \begin{equation*} y_p = A \cos x + B \sin x \text{.} \end{equation*}

   Step 2: Differentiate the guessed solution:

   \begin{gather*} y_p' = -A \sin x + B \cos x, \quad y_p'' = -A \cos x - B \sin x \end{gather*}

   Step 3: Substitute into the original equation:

   \begin{gather*} (-A \cos x - B \sin x) + (A \cos x + B \sin x) = 3 \cos x \\ 0A \sin x + 0B \cos x = 3 \cos x \end{gather*}

   Step 4: Solve for \(A \) and \(B \text{:}\)

   \begin{gather*} A = 3, \quad B = 0 \end{gather*}

   Step 5: Write the particular solution:

   \begin{equation*} y_p = 3 \cos x \text{.} \end{equation*}
6. \(\ds y'' + 3y' - 28y = 7t + e^{4t} - 1 \text{.}\)

   Solution.

   Following the method of Undetermined Coefficients, we find \(y_h\) as \(\ds y_h = c_1 e^{4t} + c_2 e^{-7t} \) and then find \(y_p\) through the steps that follow.

   Select Initial \(y_p\). The initial form of \(y_p\) is

   \begin{equation*} y_p = At + B + Ce^{4t}\text{,} \end{equation*}

   Adjust \(y_p\). Notice that \(y_p\) has an \(e^{4t}\) term, which overlaps with \(y_h\text{.}\) Therefore, we need to adjust \(y_p\) by multiplying the \(e^{4t}\) term by \(t\text{,}\)

   \begin{equation*} y_p = At + B + Cte^{4t}\text{.} \end{equation*}

   Now that \(y_p\) and \(y_h\) are independent, we can proceed to find the coefficients \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)

   Find the Coefficients of \(y_p\). Before plugging in \(y_p\text{,}\) let’s find \(y_p'\) and \(y_p''\text{:}\)

   \begin{align*} y_p =\amp\ At + B + Cte^{4t} \\ y_p' =\amp\ A + Ce^{4t} + 4Cte^{4t} \\ y_p'' =\amp\ 8Ce^{4t} + 16Cte^{4t} \text{.} \end{align*}

   Now, substituting these into the equation and collecting like terms, to get

   \begin{align*} 8Ce^{4t} + 16 Cte^{4t} \amp\ + 3(A + Ce^{4t} + 4Cte^{4t}) \\ \amp\ - 28(At + B + Cte^{4t}) = 7t + e^{4t} - 1\\ \us{1}{\ul{(11C)}}e^t + \us{2}{\ul{(-28A)}}\amp t + \us{3}{\ul{(3A-28B)}} = \us{1}{\ul{(1)}}e^{4t} + \us{2}{\ul{(7)}}t + \us{3}{\ul{(-1)}}\text{,} \end{align*}

   which leads to the following equations for \(A\text{,}\) \(B\text{,}\) and \(C\text{:}\)

   \begin{align*} \amp\ 1. \\ \amp\ 2. \\ \amp\ 3. \end{align*}

   \begin{align*} 11C =\amp\ 1 \\ -28A =\amp\ 7 \\ 3A - 28B =\amp\ -1 \end{align*}

   \begin{gather*} \\ \rightarrow \\ \end{gather*}

   \begin{align*} \amp\ C = 1/11 \\ \amp\ A = -1/4 \\ \amp\ B = 1/112 \end{align*}

   Finally, the general solution is

   \begin{equation*} y = y_h + y_p = c_1 e^{4t} + c_2 e^{-7t} - \frac{1}{4}t + \frac{1}{112} + \frac{1}{11}e^{4t} \text{.} \end{equation*}
7. \(\ds y'' - 100y = 9{t^2}{e^{10t}} + \cos t - t\sin t \text{.}\)
8. \(\ds 4y'' + y = e^{-2t}\sin \left( t/2 \right) + 6t\cos \left( t/2 \right) \text{.}\)
9. \(\ds 4y'' + 16y' + 17y = e^{-2t}\sin \left( t/2 \right) + 6t\cos \left( t/2 \right) \text{.}\)
10. \(\ds y'' + 8y' + 16y = e^{-4t} + \left( {t^2} + 5 \right)e^{-4t} \text{.}\)

Example D.17.