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What is the lowest ordered differantial equation which has the particular solution $y_p=(x^2-1)e^{-x}+x$?

For $e^{-x}$ we have $(D+1)$ and for $xe^{-x}\to(D+1)^2$, For $x^2e^{-x}\to (D+1)^3$

I dont know how to combine them. answer is $(D+1)^3D^2y=0$ btw.

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  • $\begingroup$ hint: $(x^2 - 1)e^{-x} = x^2e^{-x} - e^{-x}$ $\endgroup$
    – hunter
    Commented Jul 14, 2016 at 7:50
  • $\begingroup$ isnt it $(D+1)^3-(D+1)+D^2$? $\endgroup$
    – mayilma
    Commented Jul 14, 2016 at 8:29
  • $\begingroup$ Hint (systematic, no ingeniosity required): expand the LHS of $$\left(\frac{y_p(x)-x}{x^2-1}e^x\right)'=0$$ and cancel the factor $e^x$ to get (a differential equation similar to) $$(x^2-1)y'_p+(x^2-2x-1)y_p-x^3+x^2+x+1=0.$$ $\endgroup$
    – Did
    Commented Jul 14, 2016 at 8:39
  • $\begingroup$ could you please make an answer :( $\endgroup$
    – mayilma
    Commented Jul 15, 2016 at 7:30
  • $\begingroup$ @Did: Given that the answer provided by the OP at the end of the question is an operator of order 5, I suspect that the OP is looking for an equation with constant coefficients (even though this is not explicitly mentioned in the question). Otherwise the $0$-th order equation $y = y_p$ would do... $\endgroup$
    – Alex M.
    Commented Jul 17, 2016 at 14:12

1 Answer 1

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Given: $$y_p = (x^2-1)e^{-x}+x$$ Finding the total derivatives we obtain: $$dy = [e^{-x}(1+2x-x^2) + 1]dx$$ $$\therefore \frac{dy}{dx} = 1 - e^{-x}(x^2-2x-1)$$ $$\qquad \qquad\qquad\qquad\qquad\qquad\qquad\text{is the required differential equation} $$

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