I'm attempting to solve a Cauchy-Euler form ODE by reduction of order. I've found an alternative solution online that is correct but the exact methodology being performed isn't outlined.
Reduce the order of the following ODE and solve,
$2x^2z'''+xz''-3z'=2/x^3, x>0$
Solution:
Place the equation in Cauchy-Euler form by multiplying by $x$,
$2x^3z'''+x^2z''-3xz'=2/x^2$
We reduce the above equation by making the substitution $x=e^t$ and using differential operators to obtain,
$[2D(D-1)(D-2)+D(D-1)-3D]z=2e^{-2t}$
$[2D^3-5D^2]z=2e^{-2t}$
The homogeneous solution is found by using the roots of the auxiliary equation,
$2D^3-5D^2=0 \rightarrow2r^3-5r^2=0\rightarrow r=0,0,5/2$
$z_h=C_1e^{0t}+C_2xe^{0t}+C_3e^{5/2t}$
Now the next step would be to find the particular solution using typical method undetermined coefficients by comparing the standard linear operator $L[z]=Cx^me^{rx}$ to $g(x)=2e^{-2t}$ and finding $y_p=x^s[A_m+…A_1+A_0]e^{rx}$. From there, we'd find successive directives of $y_p$ to substitute into the original Cauchy-Euler equation to determine the coefficients of the particular solution.
However, a correct answer I found online uses the method below instead:
Consider $[2D^3-5D^2]z=2e^{-2t}$ and solve for $z_p$ as $z_p=(2e^{-2t})/[2D^3-5D^2]$. Now substitute $D=-2$ to obtain,
$z_p=(2e^{-2t})/[2(-2)^3-5(-2)^2]=-e^{-2t}/18$
Substitute $t=ln(x)$ to obtain the final particular solution of,
$z_p=-x^2/18$.
I've seen this method of solving for the particular solution used in other questions but there is never any explanation of what exact method and logic is being used. My textbook has no information either. What is this method called? Why can we simply rearrange for $z_p$ using the differential operator equation? Where does the substitution $D=-2$ come from?
Best Answer
$$2x^2z'''+xz''-3z'=\dfrac 2 {x^3}, x>0$$ Note that you can also rewrite the DE as: $$2x^2z'''\color {blue}{+4xz''-3xz''}-3z'=\dfrac 2 {x^3}$$ $$(2x^2z''-3xz')'=\dfrac 2 {x^3}$$ $$(2x(xz')'-5xz')'=\dfrac 2 {x^3}$$ You can substitute $u=xz'$. And integrate.
I suppose that $a \ne b$ for simplicity: $$y_p=\dfrac 1 {D-a}e^{bx}$$ $$(D-a)y_p=e^{bx}$$ $$y'_p-ay_p=e^{bx}$$ We seek an integrating factor. And we find that it is $\mu (x)= e^{-ax}$ mulitply both side by the integrating factor then we have: $$y'_pe^{-ax}-ae^{-ax}y_p=e^{bx}e^{-ax}$$ More simply: $$(y_pe^{-ax})'=e^{(b-a)x}$$ Now integrate both sides. We dont need the constant of integration: $$y_pe^{-ax}=\dfrac {e^{(b-a)x}}{b-a}$$ $$y_p=\dfrac {e^{bx}}{b-a}$$ We just put $D=b$ in the first line. This explains the trick. $$y_p=\dfrac {e^{bx}}{\color {red}{D-a}}=\dfrac {e^{bx}}{\color {red}{b-a}}$$