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In the above image, (Example 13) I have formed the Auxiliary equation i.e. And tried to solve the equation by "real root theorem". But could not find the solution.

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  • $\begingroup$ So the equation is $$y^{(4)}-4y'''+8y''-8y'+4y=0$$?? $\endgroup$ – Dave Nov 16 '17 at 15:52
  • $\begingroup$ Yes. It is the Question . $\endgroup$ – Akshie Dhiman Nov 16 '17 at 15:54
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HINT $$ \begin{split} p(m) &= m^4 - 4m^3 + 8m^2 - 8m + 4 \\ &= \left(m^2 - 2m + 2\right)^2 \end{split} $$

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  • $\begingroup$ Thank you so much. This is helpful. But I m having one doubt i.e. "Is real root theorem always applicable , as in this question I cannot find the zero's of this polynomial by using real root theorem $\endgroup$ – Akshie Dhiman Nov 16 '17 at 16:00
  • $\begingroup$ @AkshieDhiman see update, same idea -- you can factor into complex roots instead of real roots and end up with trigonometric solutions using De Moivre's formula $\endgroup$ – gt6989b Nov 16 '17 at 16:01
  • $\begingroup$ Okay thank you so much $\endgroup$ – Akshie Dhiman Nov 16 '17 at 16:03

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