A curve is such that dy/dx=k/x^2 +2 .

Given that the curve passes through the points (1,2) and (0,4). Find the equation of the curve.[/FONT]

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- Apr 19th 2011, 06:52 PMreniiIntergration
A curve is such that dy/dx=k/x^2 +2 .

Given that the curve passes through the points (1,2) and (0,4). Find the equation of the curve.[/FONT] - Apr 19th 2011, 06:56 PMmr fantastic
Whilst the latex compiler is down, please just post equations in the standard way. Do you mean dy/dx = 2k/x^2 ?

Integrate and use the given information "curve passes through the points (1,2) and (0,4)." to find k and the arbitrary constant of integration.

If you need more help, please show all your work and say where you are stuck. - Apr 19th 2011, 07:00 PMrenii
i have corrected it...but it is not 2k...it is dy/dx= k on x square plus 2

- Apr 19th 2011, 07:15 PMjohnny
If dy/dx = k/(x² + 2), then I got y = (k arctan(x/√2))/√2 + C.

- Apr 19th 2011, 08:53 PMTaylorM0192
The integration is correct (although maybe it's clearer to write as (k/sqrt(2)) * atan(x/sqrt(2)) + C).

Now substitute the two given points and solve the resulting system of equations:

(1). 2 = (k/sqrt(2)) * atan(1/sqrt(2)) + C

(2). 4 = (k/sqrt(2)) * atan(0) + C.

You ought to be able to determine from basic trigonometry what the exact values of the arctangent terms are; after this you will obtain a very simple set of linear equations which you can solve in a variety of ways.

NOTE: in your original problem you had (1/x^2) + 2. Had you carried out the integration and substitution of initial conditions, you would have quickly realized that the problem is impossible (i.e. there is no such curve which satisfies the conditions of the problem).

BTW, this is a basic example of solving a differential equation, so get use to this type of problem :)