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{\bf Question}

Find the particular integral for the differential equation
$$\frac{d^2x}{dt^2} + 3 \frac{dx}{dt} + 2x = 1 + 2t + t^2$$


 \vspace{.25in} {\bf Answer}

There are many Particular Integrals, but we use the method of
undetermined coefficients.

Let \begin{eqnarray*} x^* &amp; = &amp; a_0 + a_1t + a_2t^2 \\
\frac{dx^*}{dt} &amp; = &amp; a_1 + 2a_2t \\ \frac{d^2x^*}{dt^2} &amp; = &amp;
2a_2
\end{eqnarray*}

We try to fix $a_0$, $a_1$, $a_2$

Substitute into equation

$\begin{array}{rclr} \displaystyle \frac{d^2x^*}{dt^2}+
3\frac{dx^*}{dt} + 2x &amp; = &amp; 1+ 2t + t^2 &amp; (A) \\ &amp; = &amp; 2a_2 +
3(a_1 + 2a_2t)+2(a_0+a_1t+a_2t^2) \\ &amp; = &amp; [2a_2 + 3a_1 + 2a_0] +
[6a_2 + 2a_1]t + 2a_2t^2 &amp; (B)
\end{array}$

Comparing coefficients in (A) and (B)

$\begin{array}{lcl} 2a_2 + 3a_1 + 2a_0 &amp; = &amp; 1 \\ 6a_2 + 2a_1 &amp; =
&amp; 2 \\ 2a_2 &amp; = &amp; 1 \end{array}$

Solving gives $a_2 = \frac{1}{2}$, $a_1 = -\frac{1}{2}$ and $ a_0
= \frac{3}{4}$



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