Equality of two particular solutions of 2nd order linear ODE

[Harsh Bhardwaj]

I got the following two integral for the a particular solution of a 2nd order linear ODE $$(D-a)(D-b)y = g(x)$$
by using inverse operators $\frac{1}{D-a}$ and $\frac{1}{D-b}$. The two different integrals are obtained by operating these operators in different order on y to get a particular solution. It's my hunch that these integrals should be equal however I don't know how to prove it. I tried plugging in some exponential, trigonometric and power functions for g(x) and setting a and b. Both the integrals gave the exact same result(All arbitrary constants were dropped as I seek a particular solution).

$$y = e^{bx}\int e^{-bx}e^{ax}(\int e^{-ax}g(x)dx)dx$$
$$y = e^{ax}\int e^{-ax}e^{bx}(\int e^{-bx}g(x)dx)dx$$

I tried using integration by parts in different combinations and failed to prove the two integrals to be equal. Maybe the Leibniz's integral rule can be used to pull out the exponentials but I am not sure if and how it can be used for indefinite integrals.
Help me prove(or disprove) the equality of these two integrals. Thank you.

 

[Mark44]

I don't know if it's possible to evaluate the integrals for an arbitrary function g(x).

 

[Harsh Bhardwaj]

I don't know if it's possible to evaluate the integrals for an arbitrary function g(x).

I tried with some particular g(x). It was easy to prove that the two integrals are equal for g(x) = sin(kx), cos(kx) and exp(kx). I tried with g(x) = x^n but couldn't prove it(used reduction formulae and principle of induction). But all the examples I take by fixing n give the same result. Maybe I am just being lucky. It would be very helpful if someone can point out a counter example for g(x) such that the integrals are not equal and there will be no need for the proof.

 

[Erland]

If $y(x)$ is a twice differentiable function on an open interval $I$, then we easily obtain (on $I$): $(D-a)(D-b)y(x)=D^2y(x)-(a+b)\,Dy(x)+ab\,y(x)$, and likewise that $(D-b)(D-a)y(x)=D^2y(x)-(a+b)\,Dy(x)+ab\,y(x)$, so these are equal for all twice differentiable functions $y(x)$ on $I$.
Therefore, any particular solution of $(D-a)(D-b)y(x)=g(x)$ on $I$ is also a particular solution of $(D-b)(D-a)y(x)=g(x)$, and vice versa.

 

[Harsh Bhardwaj]

If $y(x)$ is a twice differentiable function on an open interval $I$, then we easily obtain (on $I$): $(D-a)(D-b)y(x)=D^2y(x)-(a+b)\,Dy(x)+ab\,y(x)$, and likewise that $(D-b)(D-a)y(x)=D^2y(x)-(a+b)\,Dy(x)+ab\,y(x)$, so these are equal for all twice differentiable functions $y(x)$ on $I$.
Therefore, any particular solution of $(D-a)(D-b)y(x)=g(x)$ on $I$ is also a particular solution of $(D-b)(D-a)y(x)=g(x)$, and vice versa.

These integrals give particular solution of y.I was wondering whether these solutions are the exact same functions. It is easy to see that the order of differentiation operators does not matter but is the same true for inverse operators? I did quite a few examples and the changing the order of inverse operators gave the exact same particular solution. My question is that is this true in general?

 

[Erland]

These integrals give particular solution of y.I was wondering whether these solutions are the exact same functions. It is easy to see that the order of differentiation operators does not matter but is the same true for inverse operators? I did quite a few examples and the changing the order of inverse operators gave the exact same particular solution. My question is that is this true in general?

Since the "inverse operator" gives the general solution, and the general solution is just the set of particular solutions, given parametrically (with two integration constants), and since the sets of particular solutions are the same in both cases, the answer is: Yes, we get the same set of solutuions in both cases.

 

[pasmith]

These integrals give particular solution of y.I was wondering whether these solutions are the exact same functions. It is easy to see that the order of differentiation operators does not matter but is the same true for inverse operators? I did quite a few examples and the changing the order of inverse operators gave the exact same particular solution. My question is that is this true in general?

General rule of function composition: $(AB)^{-1} = B^{-1}A^{-1}$. Thus if $AB = BA$ then $$B^{-1}A^{-1} = (AB)^{-1} = (BA)^{-1} = A^{-1}B^{-1}$$ as required.