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Obtain the solution for $$( D^4 + 6 D^3 + 9 D^2) y=0,$$ where $D = \frac{d}{dx}$, where $$x=0, y=0, y'=0, y''= 6,$$ and as $x$ tends to $+\infty$, $y'$ tends to $1$. For this particular solution, find the value of $y$ when $x =1$.

I solved this as follows:

The roots associated with the auxillary eqn of the given differential eqn are $m= 0,0,-3,-3$. The general solution is $$y(x) = a + b x+ c \exp(-3x) + d x \exp(-3x).$$ Then, \begin{align} y' (x) &= b - [3 c \exp (-3x) ]- [ 3 d x \exp(-3x)]+ d \exp(-3x) \\ y''(x) &= 9 c \exp(-3x)+ (9x-6) d \exp(-3x). \end{align} On substituting the initial conditions to above eqns I got \begin{align} a+c &=0\\ b-3c+d &=0\\ 9c-6d &= 6 . \end{align} Now I am not getting how to solve these for the constants $a,b,c,d$.

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    $\begingroup$ Use MathJax to write your equations .It will be more comprehensive and understandable. $\endgroup$ – SchrodingersCat Oct 22 '15 at 10:11
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I think you mean,

as $x$ tends to $+\infty$, $y'$ tends to $1$,

and not as $x$ tends to $0$.

This gives, in addition to the three (linear) conditions you've already identified, a fourth constraint, that is, $$\lim_{x \to \infty} y'(x) = b = 1.$$ These four conditions together determine the unknown coefficients $a, b, c, d$.

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  • $\begingroup$ Yes, I am extremly sorry for the mistake. $\endgroup$ – Kavita Oct 22 '15 at 10:44
  • $\begingroup$ So is it second constant b=1? $\endgroup$ – Kavita Oct 22 '15 at 10:51
  • $\begingroup$ Yes, that's exactly what this says. $\endgroup$ – Travis Oct 22 '15 at 10:58
  • $\begingroup$ Yes, I got the desired answer. Your idea helped me, thanks a lot. $\endgroup$ – Kavita Oct 22 '15 at 11:03
  • $\begingroup$ You're welcome, I'm glad you found it useful. $\endgroup$ – Travis Oct 22 '15 at 11:05

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