OMG I can't believe it.

[K.I.L.E.R]

My lecturer for maths has put up worksheets and notes up on his website.
I can't believe this.
I'm so excited.

 

[K.I.L.E.R]

Can someone please tell me if I can do this?

e^x * x.e^x = ?

 

[StefanS]

Can someone please tell me if I can do this?

e^x * x.e^x = ?

Do what with it?

 

[KimB]

Can someone please tell me if I can do this?

e^x * x.e^x = ?

Multiplied? Well, that's trivial. But usually you'd not bother with that. Since e^x and xe^x are two solutions to a particular second-order differential equation, the two above forms are typically added, not multiplied.

 

[K.I.L.E.R]

So:

e^-x * e^x * x.e^x = xe^x?

 

[KimB]

Obviously. If you're not comfortable with doing manipulations like this, consider instead:

x^-1 * x^1 * y*x^1

...or:

x^-2 * x^2 * y*x^2

Clearly the answer should be obvious for any choice of value for x, y, or the exponent. Therefore the answer should be obvious for the problem at hand.

 

[K.I.L.E.R]

Thanks.