Answer: y = 4(x² − 1)(x² − 4).
Explanation:
We need to find the equation of the curve that passes through the point (2, 0).
We start by separating the variables dy/dx and y and integrating both sides:
(2x² - 5) dy/dx = 8x(y + 9)
dy/(y + 9) = (4x/(2x² - 5)) dx
Integrating both sides:
ln|y + 9| = 2ln|2x² - 5| + C
where C is the constant of integration.
Rewriting in exponential form:
|y + 9| = e^(2ln|2x² - 5| + C)
|y + 9| = e^(ln|2x² - 5|² + C)
|y + 9| = k(2x² - 5)²
where k is the constant of integration.
Since the curve passes through the point (2, 0), we can substitute these values into the equation above to find k:
|0 + 9| = k(2(2)² - 5)²
9 = k(36)
k = 1/4
Substituting this value of k back into the equation, we get:
|y + 9| = (1/4)(2x² - 5)²
y + 9 = (1/4)(2x² - 5)² or y + 9 = -(1/4)(2x² - 5)²
Simplifying the right-hand side of each equation, we get:
y + 9 = (1/4)(4x⁴ - 20x² + 25)
or
y + 9 = -(1/4)(4x⁴ - 20x² + 25)
Expanding and simplifying, we get:
y = 4x⁴/4 - 5x²/2 + 25/4 - 9 or y = -4x⁴/4 + 5x²/2 - 25/4 - 9
y = x⁴ - 5x² + 19/4 or y = -x⁴/4 + 5x²/2 - 41/4
Thus, the equation of the curve passing through the point (2, 0) with the given gradient is y = 4(x² − 1)(x² − 4).