Answer to Question #167405 in Calculus for Rhon

Question #167405

The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve (𝑑^2𝑦) /(𝑑π‘₯^2) = 2 βˆ’ 4π‘₯. Find an equation of the curve.


1
Expert's answer
2021-03-01T07:16:16-0500

Solution

If (𝑑^2𝑦) /(𝑑π‘₯^2) = 2 βˆ’ 4π‘₯, then dy/dx = ∫(2-4x)dx = 2x-2x2+C

and y(x) = ∫(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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