Answer to Question #311008 in Differential Equations for kxngToooch

Question

Find the general solutions of the following differential equations using D-operator methods:

(D^2 - 2D + 5) y = x + 5

Answer 1

The auxiliary equation is, $m^2 -2m+5 = 0$.

Solving for m, $m = \dfrac{2\pm\sqrt{4 - 4\cdot 1 \cdot 5}}{2} = \dfrac{2\pm\sqrt{-16}}{2} = 1\pm2i$

The complementary function is, $\text{C.F} = e^x(c_1 \cos 2x + c_2 \sin 2x)$

$\begin{aligned} \text{Particular Integral} &= \frac{1}{D^2-2D+5} (x+5)\\ &=\dfrac{1}{5(1+(\frac{D^2 - 2D}{5}))}(x+5) \\ &=\dfrac{1}{5}\left(1+(\frac{D^2 - 2D}{5})\right)^{-1} (x+5)\\ &=\dfrac{1}{5}\left(1 - (\frac{D^2 - 2D}{5}) + (\frac{D^2 - 2D}{5})^2 +\cdots \right)(x+5)\\ &=\dfrac{1}{5}\left(1 - (\frac{D^2 - 2D}{5})\right)(x+5)~~~(\text{Neglecting higher differentials})\\ &=\dfrac{1}{5}\Big((x+5) - \frac{1}{5}\left(D^2(x+5) - 2D(x+5)\right)\Big)\\ &=\dfrac{1}{5}\Big(x+5 + \frac{2}{5}\Big)\\ \therefore P.I &=\dfrac{1}{25}\Big(5x+ 27\Big)\\ \end{aligned}$

The general solution is, $y= e^x(c_1 \cos2x + c_2 \sin 2x)+ \dfrac{5x+27}{25}$