Solve the differential equation dy/dx = -xe^y
and determine the equation of the slope that passes by the point P(0,1)
I get to: ∫ 1/e^y dy = - ∫ x dx
which brings me to: -1/e^y + C = -x^2/2 + C?
What do you do after?
Thank you.
so far, so good
But you don't need a +C on each side of the equation. They can be combined into a different C, since C is just an arbitrary constant. So that leaves you with
-e^-y = -x^2/2 + C
e^-y = x^2/2 + C (a different C, just - the original C)
Now take logs
-y = log(x^2/2 + C)
But that's the same as -y = log (x^2+C)/2 ( a new C)
y = log 2/(x^2+C)
Thank you very much, I get it now.
After arriving at the equation -1/e^y + C = -x^2/2 + C, you can simplify the equation by canceling out the constant C on both sides of the equation:
-1/e^y = -x^2/2
Next, you can isolate the variable y by multiplying both sides of the equation by e^y:
-1 = (-x^2/2)e^y
After multiplying both sides, you will have:
-e^y = -x^2/2
To solve for y, take the natural logarithm (ln) of both sides of the equation:
ln(-e^y) = ln(-x^2/2)
Using the property of logarithms, ln(a * b) = ln(a) + ln(b), you can rewrite the equation as:
ln(-1) + ln(e^y) = ln(-x^2/2)
Since ln(-1) is not defined, this equation has no real solutions. Therefore, there is no unique equation of the slope passing through point P(0,1) for the given differential equation dy/dx = -xe^y.