The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve (π^2π¦) /(ππ₯^2) = 2 β 4π₯. Find an equation of the curve.
Solution
If (π^2π¦) /(ππ₯^2) = 2 β 4π₯, then dy/dx = β«(2-4x)dx = 2x-2x2+C
and y(x) = β«(2x-2x2+C)dx = x2-2x3/3+Cx+D
Here C,D are arbitrary constants.
y(-1) = 1+2/3-C+D = 5/3-C+D = 3Β => D-C = 4/3
y(0) = D = 2Β Β => D = 2 and from previous equation C = 2-4/3 = 2/3
So the equation of the curve is y(x) = x2-2x3/3+2x/3+2
Answer
y(x) = x2-2x3/3+2x/3+2
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