Third order differential equation

The partial solution of homogeneous form of the differential equation is given by:

$\begin{aligned} y''' + 6y''+11y' + 6y & = 0 \\ (r+1)(r+2)(r+3)e^{rx} & = 0 \end{aligned}$

$\begin{aligned} \implies y_1(x) & = c_1 e^{-x} + c_2 e^{-2x} + c_3 e^{-3x} & \small \blue{\text{where }c_1, c_2, c_3 \text{ are constants.}} \end{aligned}$

Let the special solution be $y_2(x) = c_4 \sin 2x + c_5 \cos 2x$ , where $c_4$ and $c_5$ are constants. Then

$\begin{aligned} y_2''' + 6y_2''+11y_2' + 6y_2 & = 260 (\sin 2x + 3\cos 2x) \\ (-18c_4 - 14c_5) \sin 2x + (14c_4 - 18c_5)\cos 2x & = 260 \sin 2x + 780 \cos 2x \end{aligned}$

Equating the coefficients $\implies \begin{cases} -9c_4 - 7c_5 = 130 \\ 7c_4 - 9c_5 = 390 \end{cases} \implies \begin{cases} c_4 = 12 \\ c_5 = -34 \end{cases}$

$\implies y_2(x) = 12 \sin 2x - 34 \cos 2x$ and the general solution:

$\begin{aligned} y(x) & = c_1 e^{-x} + c_2 e^{-2x} + c_3 e^{-3x} + 12 \sin 2x - 34 \cos 2x \\ y(0) & = c_1+c_2+c_3 - 34 = -33 & \small \blue{\implies c_1+c_2 + c_3 = 1} \\ y'(0) & = -c_1-2c_2-3c_3 + 24 = 26 & \small \blue{\implies c_1+2c_2 + 3c_3 = -2} \\ y''(0) & = c_1+4c_2+9c_3 + 136 = 142 & \small \blue{\implies c_1+4c_2 + 9c_3 = 6} \end{aligned}$

Solving $\begin{cases} c_1+c_2 + c_3 = 1 \\ c_1+2c_2 + 3c_3 = -2 \\ c_1+4c_2 + 9c_3 = 6 \end{cases} \implies \begin{cases} c_1 = 11 \\ c_2 = - 17 \\ c_3 = 7\end{cases}$

$\begin{aligned} \implies y(x) & = 11 e^{-x} -17 e^{-2x} + 7 e^{-3x} + 12 \sin 2x - 34 \cos 2x \\ y\left(\frac \pi 2\right) & = 11 e^{-\frac \pi 2} -17 e^{-\pi} + 7 e^{-\frac {2\pi}2} + 34 \end{aligned}$

Therefore $A+B+C+D = 11-17+7 + 34 = \boxed{35}$ .

Solution of the given differential equation is $y=12\sin (2t)-34\cos (2t)+Ae^{-t}+Be^{-2t}+Ce^{-3t}$ . Substituting the initial conditions we get, $A=11, B=-17, C=7$ . Hence $y(\dfrac{π}{2})=34+11e^{-\dfrac{π}{2}}-17e^{-π}+7e^{-\dfrac{3π}{2}}$ . Therefore $D=34$ and $A+B+C+D=11-17+7+34=\boxed {35}$ .