How Do I Integrate Factors in Round 2 to Achieve the Proper Form for Processing?

ssb
Messages
119
Reaction score
0

Homework Statement



y' (1+e^t) + e^ty = 0

How do I get it in the form

y' +f(t)y = f(p)

That is how do apply algebra to this so it is in the proper form to process it?

The Attempt at a Solution



Kind of hard to post my attempt. I can move the y' (1+e^2) to the right side then divide both sides by y' but that doesn't make it y' + something, rather y' x something.

I also tried moving everything to the right and canceling some stuff out but it doesn't seem to work.
 
Last edited:
Physics news on Phys.org
Divide both sides by (1+e^t). Then read off f(t). What's f(p)?
 
Try dividing throughout by (1+e^t).
 
Dick said:
Divide both sides by (1+e^t). Then read off f(t). What's f(p)?

f(p) becomes

(e^ty)/(1+e^t)

giving equation


y' + (e^ty)/(1+e^t) = 0

Are you saying it is allowed to do this step even though the right hand side is zero? I always thought this was a no-no.
 
You aren't dividing by the RHS. You're dividing by (1+e^t). I'm still bothered by what 'p' is supposed to be. You've got f(t)=e^t/(1+e^t) alright.
 
Dick said:
You aren't dividing by the RHS. You're dividing by (1+e^t). I'm still bothered by what 'p' is supposed to be. You've got f(t)=e^t/(1+e^t) alright.

Gotcha. Thankyou so much. I got messed up a few weeks ago when I was given an assignment by my teacher with the wrong answer key. This is now going well. Thankyou thankyou! Thankyou
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top