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How to find the integrating factor?

I'm stuck on part b because I'm not sure how to get the integrating factor for this equation?
Screen Shot 2017-04-11 at 12.23.50.png
Reply 1
Avatar for Notnek
Notnek
21
Original post by 1974931
I'm stuck on part b because I'm not sure how to get the integrating factor for this equation?
Screen Shot 2017-04-11 at 12.23.50.png

You don't need an integrating factor - the DE is separable. Always check if a DE is separable before thinking about longer methods.
Original post by notnek
You don't need an integrating factor - the DE is separable. Always check if a DE is separable before thinking about longer methods.


oh i see thanks
wherever you see a y sub in u-x, and to find du/dx differentiate with respect to x. This will give you the equation, then seperate.
How do you even solve the first part of this?

I tried it, but this is all I got.

dydx+ey−x=1\frac{dy}{dx} + e^{y-x} = 1

u=y−xu=y-x
→dudy=1\rightarrow \frac{du}{dy} = 1
→du=dy\rightarrow du = dy
→dudx+eu=1\rightarrow \frac{du}{dx} + e^{u} = 1

I'm guessing I have to differentiate the right side with respect to u, but I don't see why
Reply 5
Avatar for B_9710
B_9710
18
Original post by hoperidesalone
How do you even solve the first part of this?

I tried it, but this is all I got.

dydx+ey−x=1\frac{dy}{dx} + e^{y-x} = 1

u=y−xu=y-x
→dudy=1\rightarrow \frac{du}{dy} = 1
→du=dy\rightarrow du = dy
→dudx+eu=1\rightarrow \frac{du}{dx} + e^{u} = 1

I'm guessing I have to differentiate the right side with respect to u, but I don't see why


dy/dx=du/dx + 1.
Original post by hoperidesalone
How do you even solve the first part of this?

I tried it, but this is all I got.

dydx+ey−x=1\frac{dy}{dx} + e^{y-x} = 1

u=y−xu=y-x
→dudy=1\rightarrow \frac{du}{dy} = 1
→du=dy\rightarrow du = dy
→dudx+eu=1\rightarrow \frac{du}{dx} + e^{u} = 1

I'm guessing I have to differentiate the right side with respect to u, but I don't see why

y-x=u
differentiate implicitly with respect to x
dy/dx -1 = du/dx
so dy/dx = du/dx +1
so
du/dx +1 +e^u=1
so du/dx +e^u=0

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