Answer to Question #311008 in Differential Equations for kxngToooch

Question #311008

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

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



1
Expert's answer
2022-03-14T18:34:36-0400

The auxiliary equation is, m22m+5=0m^2 -2m+5 = 0.


Solving for m, m=2±44152=2±162=1±2im = \dfrac{2\pm\sqrt{4 - 4\cdot 1 \cdot 5}}{2} = \dfrac{2\pm\sqrt{-16}}{2} = 1\pm2i


The complementary function is, C.F=ex(c1cos2x+c2sin2x)\text{C.F} = e^x(c_1 \cos 2x + c_2 \sin 2x)


Particular Integral=1D22D+5(x+5)=15(1+(D22D5))(x+5)=15(1+(D22D5))1(x+5)=15(1(D22D5)+(D22D5)2+)(x+5)=15(1(D22D5))(x+5)   (Neglecting higher differentials)=15((x+5)15(D2(x+5)2D(x+5)))=15(x+5+25)P.I=125(5x+27)\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=ex(c1cos2x+c2sin2x)+5x+2725y= e^x(c_1 \cos2x + c_2 \sin 2x)+ \dfrac{5x+27}{25}


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