Question

Differential equation help
y'= y + e^(x)

Answer 1

I know I have to use an integrating factor but not quite sure cause I dont really remember how to

Answer 2

The equation is linear:
\[y'-y=e^x\]
The integrating factor is \(\mu(x)=e^{-\int~dx}=e^{-x}\), so you have
\[e^{-x}y'-e^{-x}y=1~~\iff~~\frac{d}{dx}[e^{-x}y]=1\]

Answer 3

so integration factor is -1?

Answer 4

\[e^{-x}y'-e^{-x}y \to \frac{ d }{dx }[e^{-x}y]\]

Answer 5

@SithsAndGiggles Im kinda confused on seeing how you got from there to there

Answer 6

No, the integrating factor is " e^(the integral of -1 dx)

Answer 7

I didn't mean to put "No" lol. In simple terms...you multiply "everything" by the "integrating factor, integrate both sides, and reagange for y=...

Answer 8

Ok ok now but once you simpify it how does it get to\[e^{-x}y' - e^{-x}y \to \frac{ d }{ dx }e^{-x}y\]

Answer 9

The trick is that the left side of the equation is the product rule
(e^-x)(dy/dx)-e^-x *y= e^-x *y

Answer 10

I feel like it sounds like a dumb question but i have no clue how it's getting to there

Answer 11

((e^-x)(y))'=the left side

Answer 12

Get it? What is the derivative of e^-x times y?? The left side! :)

Answer 13

So once you "integrate" both sides, the integral of the left side of the equation is simply itself, because it is a "derivative"