# $(D^5-D)y = 12e^x$ using exponential shift

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?

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 $$3xe^x$$.
• $$12e^x \stackrel{(D^4+D^3+D^2+D)^{-1}}\Longrightarrow 3e^x\stackrel{(D-1)^{-1}}\Longrightarrow 3xe^x.$$ $$12e^x \stackrel{D^4+D^3+D^2+D}\Longleftarrow 3e^x\stackrel{D-1}\Longleftarrow 3xe^x.$$Can you please be more specific about which part of the argument you don't understand? – Song Jan 19 at 16:42
• @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 $(D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x$. – Song Jan 19 at 17:41