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The question is to find the particular integral of the expression $$ (D^3+1)y = cos(2x-1) $$I’m not sure how to go about it. How do I find the PI for this expression?

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2 Answers

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$\require{extpfeil}\Newextarrow{\xRightarrow}{5,5}{0x21D2}$This is an interesting differential equation because it yields some really nice graphs when some initial $y(0),y'(0),y''(0)$ are sampled.

As for the solution, first solve the homogenous problem:$$y''' + y = 0 \xRightarrow{y = e^\lambda} \lambda^3 e^\lambda + e^\lambda = 0 \Rightarrow \lambda = \begin{cases} -1, \\ \dfrac{1}{2} + \dfrac{i\sqrt{3}}{2}, \\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2}.\end{cases}$$

Thus, the general solution is:\begin{align*} y_g(x) &= c_1e^{-x} + c_2 e^{x\left(\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)} + c_3 e^{x\left(\frac{1}{2} - \frac{i\sqrt{3}}{2}\right)} \\ &= c_1 e^{-x} + c_2 e^{x/2}\sin\left(\frac{\sqrt{3}}{2}x\right) + c_3e^{x/2} \cos\left(\frac{\sqrt{3}}{2}x\right). \end{align*}

For the particular solution (aka your particular integral), based on the expression of the given differential, as Ninad Munshi mentioned in the comments, one can "guess" the expression:

$$y_p(x) = PI = A\cos(2x-1) + B\sin(2x-1).$$Substitute then in the initial expression and calculate the values of $A$ and $B$.

Finally, the interesting part that I promised:

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For the particular integral it's more simple to change the variable first $u=2x-1$$$(8D^3+1)y = \cos(u)$$$$y_p= \dfrac {\cos(u)}{(8D^3+1)} $$$$y_p= \dfrac {\cos(u)}{(-8D+1)} $$$$y_p= \dfrac {8D+1}{(-64D^2+1)} \cos u $$$$y_p= \dfrac {8D+1}{65} \cos u $$$$y_p= \dfrac {-8 \sin u}{65} +\dfrac {\cos u }{65}$$Unsubstitute $u=2x-1$$$\boxed {y_p= \dfrac {-8 \sin (2x-1)}{65} +\dfrac {\cos (2x-1) }{65}}$$

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