- #1
Harsh Bhardwaj
- 17
- 3
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.
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.