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Find the particular solution using the method of undetermined coefficients:

$$y''''+y'''=1-x^2e^{-x}.$$

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Your equation is $y''''+y'''=1-x^2e^{-x}$. Now change the $y'$ to $Dy$ form as follows. So $$y''''\to D^4y,\\ y'''\to D^3y,$$ So by new arranging respect to $D$ operator we get our equation as: $$D^4y+D^3y=1-x^2e^{-x}$$ or by factoring and expanding $$(D^4+D^3)y=D^3(D+1)y=1-x^2e^{-x}$$ Note that considering the corresponding homogeneous equation $$D^3(D+1)y=0$$ we get $D^3=0,\;\; (D+1)=0$ which leads us to write the general solutions as $$y_c(x) = c_1+c_2x+c_3x^2+c_4e^{-x}$$ But some facts:

  • If $y=\text{constant}$ so $y'=0$ or $Dy=0$. Here, the operator $D$ annihilates $y$ which is just a constant.($Dc=0$)
  • If $y=cx$ in which $c$ is a constant so $y''=0$ or $D^2y=0$. It means that the operator polynomial $P(D)=D^2$ annihilates $y=cx$. $(\text{or} \; P(D)=D^2(cx)=cD^2x=c(x'')=0)$. Generally, $D^{n+1}$ annihilates not only the function $y=cx^{n}$ but also all linear functions as $$y=c_0+c_1x+c_2x^2+...+c_nx^n$$ It means that $$P(D)y=D^{n+1}y=0$$.
  • As the same the differential operator $(D-\alpha)^n$ annihilates each of the following functions and every linear combinations of them: $$e^{\alpha x},xe^{\alpha x},x^2e^{\alpha x},...,x^{n-1}e^{\alpha x}$$ Now look at the RHS of your original equation. I mean $=1-x^2e^{-x}$. Can we guess of which proper differential polynomial annihilates it? As above it would be $D(D+1)^3$. $\;$ $D$ for $1$ and $(D+1)^3$ for $-x^2e^{-x}$. It means that $D(D+1)^3 \left(1-x^2e^{-x}\right)=0$.

Now we consider of what we have achieved at last: $$D^3(D+1)y=1-x^2e^{-x}$$ Put the operator $D(D+1)^3$ before both sides of the above converted equation: $$D(D+1)^3\left(D^3(D+1)y\right)=D(D+1)^3 \left(1-x^2e^{-x}\right)=0$$ or $$D^4(D+1)^4y=0$$ In fact, we have found a proper differential operator $P(D)=D^4(D+1)^4$ which if it effects to $y$, $y$ will be lost.

Now, for a while, forget our equation and look at $D^4(D+1)^4y=0$ and think someone gave this to us asking to guess which function $y$ may satisfy the equality above? We reply:

  • Since we have $(D+1)$ so we have some forms as $e^{-x}$ in $y$.
  • Since $(D+1)$ has a power $4$ so we have the forms $Ae^{-x},\; Bxe^{-x}, \;Cx^2e^{-x}, \; Ex^3e^{-x}$ in $y$. Note that you multiply $e^{-x}$, in the previous line, by $A,\; Bx,\; Cx^2,\; Ex^3$. (Exactly until the power of $x$ becomes $4-1=3$).

  • And, since we have $D^4$, then $y$ has the term $F,\; Gx,\; Hx^2,\; Lx^3$.

So we are done. Our probable function which satisfy the original equation is $$y=Ae^{-x}+ Bxe^{-x}+Cx^2e^{-x}+Ex^3e^{-x}+F+Gx+Hx^2+Lx^3$$. Now eliminate those terms which generate $y_c(x)$ above and take the rest for what we have looked for. After finding the $y_p(x)$, you should satisfy it into the equation for finding the unknown coefficients.

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