A-LevelMathematicsDifferential equations9709/32 · May/June 2021 · Q77 Marks

A curve is such that the gradient at a general point with coordinates (x, y) is proportional to y / √(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.

A curve is such that the gradient at a general point with coordinates (x, y) is proportional to y / √(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.[7]

Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[B1] State equation dy/dx = k y/√(x+1)
[M1] Separate variables correctly for their differential equation and integrate at least one side
[A1] Obtain ln y
[A1] Obtain 2[k]√(x+1)
[DM1] Use (0, 1) and (3, e) in an expression containing ln y and √(x+1) and a constant of integration to determine k and/or a constant of integration c (or use (0, 1), (3, e) and (x, y) as limits on definite integrals)
[A1] Obtain k = 1/2 and c = −1
[A1] Obtain y = exp(√(x+1)−1)

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