Question

Algebra problem

Answer 1

Just looking at a solution I was given by my teacher. Is there an error at the end

Answer 2

Looks like he's dividing the expontial but I don't know how he's getting A*e^-4x

Answer 3

e^-4x cancels on the fractions but not the constant for some reason.

Answer 4

no problem with that...just multiply by \(e^{4x}\) to get u

Answer 5

@mukushla So he's just multiplying every term e^-4x?

Answer 6

If so, where are the e^-4x on the fractions??

Answer 7

Well your teacher multiplied the expression by e^4x. This cancels out the e^-4x on the constants leaving e^4x only on the A1 term.Got it?

Answer 8

It makes no sense to me.

Answer 9

The expontials magically dissppear

Answer 10

@UnkleRhaukus any ideas?

Answer 11

Generally (x^a) * (x^-a) = x^ ( a+(-a)) = 1. That property is used here.

Answer 12

Here's the solution, in it's entire form, a lot of computation in there.

Answer 13

\[ue^{4x}=-\frac{xe^{-4x}}{4}-\frac 1{16}e^{-4x}+A_1\]
\[u=-\frac{xe^{-4x}}{4}e^{4x}-\frac 1{16}e^{-4x}e^{4x}+A_1e^{4x}\]
\[u=-\frac{x}{4}-\frac 1{16}+A_1e^{4x}\]

Answer 14

i think i might have the first term wrong ,

Answer 15

So are dividing across using that exponential. Why divide by the A*e....

Answer 16

I'll ask a few more to come in any ideas @sami-21

Answer 17

your teacher looks like hes equating u with the original that has e^-4x factored out of it