@@giuseppemalaguti435 (x+1)e^(x+x^2/2)=-1 Integrating from -1 to some X we get: e^(x+x^2/2)= -(x+1) But we know that LHS is also equal to -1/(x+1), thus: -(x+1) = -(1/x+1) (x+1)^2 = 1 => x = -2 or x = 0 I wonder if it's a coincidence
I am always confused how solving for solutions and then you check to find some 'don't work' Any approach in these "Lambert's" problems to only find working solutions?
exp (( ( x + 1) ^2 - 1) /2) = - 1/( x + 1)
exp ( (z^2 - 1) /2) = - 1/z
Has a trivial solution at x + 1 = z = -1
This implies x = - 2
Well done
Clearly -2 is the only solution. The range values set tne value of x to be -2. Archimides decomposition wiill work as well.Equations everyday.yeah..
Very nice problem and solution.
Thank you 🙂
We should have expected that after squaring both sides we could have introduced extraneous solutions. That's what happened with x=0!
He mentioned it.
Cool
Loved it
Nice!
It's nice 👍👍👍
Can you multiply by (x+1) and integrate both sides? LHS is f'e^f
??? Non si può fare
@@giuseppemalaguti435
(x+1)e^(x+x^2/2)=-1
Integrating from -1 to some X we get:
e^(x+x^2/2)= -(x+1)
But we know that LHS is also equal to -1/(x+1), thus:
-(x+1) = -(1/x+1)
(x+1)^2 = 1
=> x = -2 or x = 0
I wonder if it's a coincidence
👍
x=-√W(e)-1=-2
I am always confused how solving for solutions and then you check to find some 'don't work'
Any approach in these "Lambert's" problems to only find working solutions?
It's wrong from about 4 minutes
what’s wrong
Sorry I was wrong