Given \(\frac{dy}{dx} = \sqrt{2x + 5}\), we need to find \(y\) by integrating.
Integrate \(\frac{dy}{dx} = \sqrt{2x + 5} = (2x + 5)^{1/2}\).
\(y = \int (2x + 5)^{1/2} \, dx\)
Using the substitution \(u = 2x + 5\), \(du = 2 \, dx\), so \(dx = \frac{1}{2} \, du\).
\(y = \int (u)^{1/2} \cdot \frac{1}{2} \, du\)
\(y = \frac{1}{2} \int u^{1/2} \, du\)
\(y = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + c\)
\(y = \frac{1}{3} (2x + 5)^{3/2} + c\)
Given the point \((2, 5)\) is on the curve, substitute \(x = 2\) and \(y = 5\):
\(5 = \frac{1}{3} (2(2) + 5)^{3/2} + c\)
\(5 = \frac{1}{3} (4 + 5)^{3/2} + c\)
\(5 = \frac{1}{3} (9)^{3/2} + c\)
\(5 = \frac{1}{3} \cdot 27 + c\)
\(5 = 9 + c\)
\(c = -4\)
Thus, the equation of the curve is:
\(y = \frac{1}{3} (2x + 5)^{3/2} - 4\)
\(y = \frac{2}{3} (2x + 5)^{3/2} - 4\)
