A weird differential equation

The fraction on the LHS represents both a differential operator and an inverse differential operator. Note that for differential operators,

$(1-D)(1+D+D^2+D^3+...)f(x) = f(x)$

which, surprisingly, aligns with the properties of an infinite geometric series.

We can follow from the relation above and get this

$(1-D^2)(1+D^2+D^4+D^6+...)f(x)=f(x)$

and find that

$(1-D^2)f(x) = (\frac{1}{1+D^2+D^4+D^6+...}) f(x)$

So now the differential equation reduces to

$D(1-D^2)y= e^{\frac{1}{2}x} + e^{\frac{1}{3}x} + x^4$

This enables us to directly find the complementary solutions immediately (knowing that the eigenvalues of the characteristic polynomial are $0$ , $1$ , and $-1$ ). So now we have the complementary solution

$y_c = c_1 + c_2e^x + c_3e^{-x}$

Now, for the particular solution, we can find it using method of undetermined coefficients, variation of parameters, or by annihilators. But all will prove too tedious and lengthy. Let's just go back to the original equation.

$\frac{D}{1+D^2+D^4+D^6+...} \{y\} = e^{\frac{1}{2}x} + e^{\frac{1}{3}x} + x^4$

for any differential equation with a differential operator $L(D) y = r(x)$ , the particular solution would be $y = L^{-1} r(x)$ .

In such case, we take the operator $(1+D^2+D^4+D^6+...)$ and operate it on both sides, resulting to

$Dy = (1+D^2+D^4+D^6+...) (e^{\frac{1}{2}x} + e^{\frac{1}{3}x} + x^4)$

Now, let us evaluate each term in the right.

Note that for polynomials, the infinite differential operator eventually terminates. in this case, it terminates after three terms, and that will give us $x^4 + 12x^2 + 24$ .

Now, for the exponentials, take a look at this.

$g(x) = e^{\frac{1}{2}x}$

$D^2 g(x) = \frac{1}{4} e^{\frac{1}{2}x}$

$D^4 g(x) = \frac{1}{16} e^{\frac{1}{2}x}$

This is an infinite geometric series with common ratio $\frac{1}{4}$ , beginning at $g(x)$ . Its sum would then be

$(1+D^2+D^4+...)g(x) = \frac{g(x)}{1- \frac{1}{4}} = \frac{4}{3} e^{\frac{1}{2}x}$

In a similar way, we can find the particular solution for $e^{\frac{1}{3}x}$ , and that is $\frac{9}{8}e^{\frac{1}{3}x}$ .

So now, we arrive at this.

$Dy = \frac{4}{3} e^{\frac{1}{2}x} + \frac{9}{8}e^{\frac{1}{3}x} + x^4 + 12x^2 + 24$

operating $D^{-1}$ on both sides, we finally get the particular solution as

$y_p = \frac{8}{3} e^{\frac{1}{2}x} + \frac{27}{8}e^{\frac{1}{3}x} + \frac{x^5}{5} + 4x^3 + 24x + C$

since $y = y_c + y_p$ , $C$ will be absorbed by $c_1$ , and we now get the general solution as

$\large\boxed{ y= c + c_2e^x + c_3e^{-x} + \frac{8}{3} e^{\frac{1}{2}x} + \frac{27}{8}e^{\frac{1}{3}x} + \frac{x^5}{5} + 4x^3 + 24x }$