Let f: R → R be a differentiable function with f(0) = 0. If y = f(x) satisfies the differential equation dy/dx = (2 + 5y)(5y⁻¹), then the value of lim(x→-∞) f(x) is k. Then what is the value of 5k?

dy/dx = 25y⁻¹ + 5
So, dy/(25y⁻¹ + 5) = dx
Integrating,
1/25 ∫ 1/(y⁻¹ + 1/5) dy = x + c
1/25 ∫ 5y/(5y+1) dy = x + c
1/5 ∫ (1 - 1/(5y+1)) dy = x + c
1/5[ y - (1/5)ln|5y+1| ] = x + c
y - (1/25)ln|5y+1| = 5(x+c)
Now, c = 0 as f(0) = 0
Hence, y - (1/25)ln|5y+1| = 5x
Let's consider the limit as x approaches -∞:
lim(x→-∞) [ y - (1/25)ln|5y+1| ] = lim(x→-∞) 5x
lim(x→-∞) y - (1/25)ln|5y+1| = -∞
This implies that lim(x→-∞) y = -∞, since the logarithmic term is significantly slower growing than y as x→-∞
Let's use the given equation:
dy/dx = (2 + 5y)(1/5y)
This simplifies to:
dy/dx = (2/5y) + 1
Separating variables:
dy/(1 + 2/(5y)) = dx
Integrating both sides is difficult to solve analytically, and the above steps provide a more feasible approach.
Let's reconsider the equation obtained earlier:
y - (1/25)ln|5y + 1| = 5x
As x → -∞, we assume y → k (some constant)
Then k - (1/25)ln|5k + 1| = -∞
This is only possible if k = -∞. However, this result may seem contradictory.
Let's analyze the original differential equation again:
dy/dx = (2 + 5y)(1/5y)
Let's assume lim(x→-∞) f(x) = k.
Then, as x → -∞, dy/dx → 0
(2 + 5k)(1/5k) = 0
2 + 5k = 0
k = -2/5
Therefore, 5k = 5(-2/5) = -2