Can an Exponential Factor Create a New Solution for a Second Order ODE?

[intervoxel]

Is it possible to form a new solution of a second order ODE by multiplying it by an exponential factor?

 

[g_edgar]

Not clear what you mean. In a second-order LINEAR diff eq, if you have one solution u(x) you can find the general solution by trying the form f(x)u(x) ... This should be in standard ODE textbooks.

 

[intervoxel]

What I mean is that I have two particular solutions even and odd that diverge at infinity, but I noted that if I multiply them by exp(-z^2 / 4) they behave properly. I'm trying to justify this procedure. Substituting the product of each back into the differential equation doesn't seem to work.

 

[HallsofIvy]

In general, no. Multiplying a solution to a d.e. by another function, exponential or not, does NOT give a new solution.

 

[blue_raver22]

Yes, it is possible to form a new solution of a second order ODE by multiplying it by an exponential factor. This is known as the method of variation of parameters. By multiplying the original solution of the ODE by an exponential factor, we can obtain a new solution that satisfies the same ODE. This method is commonly used in solving linear ODEs with constant coefficients. The exponential factor introduces a new parameter that allows for a more general solution to the ODE. This technique is widely used in many scientific fields, including physics, engineering, and mathematics.