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{\bf Question}

Find the general solution of the differential equation:
$$\frac{d^2x}{dt^2} + 4x = 1 + \sin 2t $$

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

Need general solution of $$\frac{d^2x}{dt^2} + 4x = 1 + \sin 2t $$

Complementary Function: $m^2 + 4 = 0 \Rightarrow  m = \pm 2j$

Hence the general solution is $\Rightarrow x_c(t) = A \cos 2t + B
\sin 2t$

For the particular integral for $\displaystyle
\frac{d^2x_1^*}{dt^2} + 4x_1 ^*  =  1$

Try  $\displaystyle x_1^*  =  c \Rightarrow \frac{d^2x_1^*}{dt^2}
= 0 \Rightarrow 4c = 1 \Rightarrow c = \frac{1}{4}$


Find a particular integral for $\displaystyle
\frac{d^2x_2^*}{dt^2} + 4x_2^* = \sin 2t$

We can not try the obvious $x_2^* = P \cos 2t +  Q \sin 2t $
because this is the same as the complementary function.  So
instead we will try

$x_2^* = t(P \cos 2t +  Q \sin 2t) $

then we shall try to find P and Q

\begin{eqnarray*}  \frac{dx_2^*}{dt} & = & (P \cos 2t +  Q \sin 2t) +
t(-2P \sin 2t +  2Q \cos 2t) \\ & = & (2Qt + P) \cos 2t + (-2Pt +
Q) \sin 2t \\  \frac{d^2x_2^*}{dt^2} & = & (-4Pt + 2Q + 2Q) \cos
2t + (-4Qt -2P -2P) \sin 2t \\ \frac{d^2x_2^*}{dt^2} + 4 x_2^* & =
& 4 \{ \cos 2t (Q -Pt + Pt) + \sin 2t (-Qt - P + Qt) \} \\ &
\equiv & \sin 2t
\end{eqnarray*}

Hence $Q = 0$ and $-4P = 1 \Rightarrow P = -\frac{1}{4}$

Hence$x_2^* = -\frac{1}{4}t \cos 2t$

The General Solution is: \begin{eqnarray*} x & = & x_c + x_1^* +
x_2^* \\ & = & \left( A - \frac{1}{4} t \right) \cos 2t + B \sin
2t + \frac{1}{4}
\end{eqnarray*}



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