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The equation is $$(D^5-D)y = 12e^x$$

Here, the general solution of homogenous (D^5-D)y=0 is $y_g = c_1 + c_2 \cos x + c_3 \sin x + c_4 e^x + c_5 e^{-x}$. Now I can apply two methods:

  1. Assume particular solution to be $Axe^x$ and solve. This is linearly independant from $y_g$. But finding $D^5$ can be hectic.

  2. Use exponential shift:

$$(D^5-D)y = 12e^x\\ y = 12 e^x\frac{1}{(D+1)^5-(D+1)}(1) \\ = 12 e^x\frac{1}{(D+1)^4-1}(1)\\ = 12e^x (1+(D+1)^4 + ....)$$

This gives strange answer. What is the proper method for this?

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1. More efficient way is as follows. Let us write $T=D^5-D$. For any $r\in\Bbb C$, we have $$ T[e^{rx}]=\frac{d^5}{dx^5}e^{rx}-\frac{d}{dx}e^{rx}=(r^5-r)e^{rx}. $$ Now, differentiate with respect to $r$. Then we get $$ \frac{\partial }{\partial r}T[e^{rx}]=T\left[\frac{\partial }{\partial r}e^{rx}\right]=T[xe^{rx}]=\frac{\partial }{\partial r}\left((r^5-r)e^{rx}\right)=(5r^4-1)e^{rx}+(r^5-r)xe^{rx}. $$ We get by letting $r=1$ $$ T[xe^x]=4e^x. $$ Hence, $T[3xe^x]=12e^x$ and $3xe^x$ is a particular solution of the equation.

2. In fact, correct formula should be $$ y=12(D^5-D)^{-1}[e^{x}], $$ not $$ y=12e^x (D^5-D)^{-1}[1]. $$ Note that $D^5-D=(D-1)(D^4+D^3+D^2+D)$. Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have $$ (D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x. $$ Finally, since $(D-1)[xe^x]=e^x$, we have $$ (D-1)^{-1}[3e^x]=(D-1)^{-1}(D^4+D^3+D^2+D)^{-1}[12e^x]=3xe^x. $$ This also gives a particular solution $3xe^x$.

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  • $\begingroup$ sorry i dont understand your method 2 $\endgroup$ – jeea Jan 19 at 16:38
  • $\begingroup$ $$12e^x \stackrel{(D^4+D^3+D^2+D)^{-1}}\Longrightarrow 3e^x\stackrel{(D-1)^{-1}}\Longrightarrow 3xe^x.$$ $$12e^x \stackrel{D^4+D^3+D^2+D}\Longleftarrow 3e^x\stackrel{D-1}\Longleftarrow 3xe^x.$$Can you please be more specific about which part of the argument you don't understand? $\endgroup$ – Song Jan 19 at 16:42
  • $\begingroup$ Yes the first line only, from 12ex to 3ex $\endgroup$ – jeea Jan 19 at 17:27
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    $\begingroup$ @jeea Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have $(D^4+D^3+D^2+D)^{-1}[4e^x]=e^x$. By multiplying $3$, we have $(D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x$. $\endgroup$ – Song Jan 19 at 17:41
  • $\begingroup$ Oh so we have to observe and predict very carefullly $\endgroup$ – jeea Jan 20 at 4:20

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