Solve Ly=y''(x)+4xy'(x)-2x for Linear Functionals

[UrbanXrisis]

I'm not quite sure if this is a linear functional but the question asks:

if $$L=D^2+4xD-2x$$ and $$y(x)=2x-4e^{5x}$$

I am to find Ly=?

My first impressions to solve this is the take $$Ly=y''(x)+4xy'(x)-2x$$

i'm not quite sure how to solve this but I got:

$$y''(x)=-100e^{5x}$$
$$y'(x)=-20e^{5x}+2$$

and then I plug it into $$Ly=y''(x)+4xy'(x)-2x$$

I don't think I did this correctly, could someone help me out?

 

[daveb]

Why do you think you did it incorrectly? It certainly looks OK to me (unless you plugged in the values incorrectly). To see whether it's a linear functional, just apply the definition of linear functional.

 

[UrbanXrisis]

I got: $$Ly=-100e^{5x}+4x(-20e^{5x}+2)-2x$$

would that be it? I plugged this into my internet answer recorder and it gave me an incorrect.. not sure why

 

[George Jones]

There is a small mistake.

The last term in L hasn't been applied correctly.

Regards,
George

 

[daveb]

oops! yeah...missed that one!

 

[UrbanXrisis]

i don't understand, what is wrong with the last term? it doesn't have a D so doesn't it stay as -2x?

 

[George Jones]

i don't understand, what is wrong with the last term? it doesn't have a D so doesn't it stay as -2x?

Remember, you're applying L to y, i.e., you're finding Ly.

Regards,
George

 

[Nimz]

If you don't see the D, that doesn't mean it isn't there (there are many examples of things not seen that are still there, and I'm sure you can come up with several). You can think of it as D⁰, if that helps.

 

[George Jones]

If L = -2x, then Ly = ?

Regards,
George

 

[UrbanXrisis]

$$-2x(2x-4e^{5x})$$

thank you for the help!