A curve is such that the gradient at a general point with coordinates \((x, y)\) is proportional to \(\frac{y}{\sqrt{x+1}}\).
The curve passes through the points with coordinates \((0, 1)\) and \((3, e)\).
By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\).
Solution
The gradient of the curve is given by \(\frac{dy}{dx} = k \frac{y}{\sqrt{x+1}}\), where \(k\) is a constant of proportionality.
Separate variables: \(\frac{1}{y} dy = \frac{k}{\sqrt{x+1}} dx\).
Integrate both sides: \(\int \frac{1}{y} dy = \int \frac{k}{\sqrt{x+1}} dx\).
This gives \(\ln y = 2k \sqrt{x+1} + c\), where \(c\) is the constant of integration.
Using the point \((0, 1)\), substitute into the equation: \(\ln 1 = 2k \sqrt{0+1} + c\), which simplifies to \(c = -2k\).
Using the point \((3, e)\), substitute into the equation: \(\ln e = 2k \sqrt{3+1} - 2k\), which simplifies to \(1 = 4k - 2k\).
Solve for \(k\): \(k = \frac{1}{2}\).
Substitute \(k = \frac{1}{2}\) and \(c = -1\) back into the equation: \(\ln y = \sqrt{x+1} - 1\).
Exponentiate both sides to solve for \(y\): \(y = \exp(\sqrt{x+1} - 1)\).
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