Find the particular solution using the method of undetermined coefficients:
$y''''+y'''=1-x^2e^{-x}.$
Find the particular solution using the method of undetermined coefficients:
$y''''+y'''=1-x^2e^{-x}.$
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:
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 $(D+1)$ 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.