Calulus question that is making little sense
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Calulus question that is making little sense

[From: ] [author: ] [Date: 12-01-02] [Hit: ]
-A) plug in values for x and y, given 12 points.Example if (1,2) is a point,at (1,(hint:depending on how exact you need to be,......
Consider the differential equation dy/dx= x^2(2y+1).

A) Sketch a slope field for the given differential equation at the 12 points provided.

You can't really show me but how can I go about doing this?

B) describe all points in the xy plane for which the slopes are positive.

C) Find the particular solution y=f(x) to the given differential equation with the initial condition f(0)=5.

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A) plug in values for x and y, given 12 points.

Example if (1,2) is a point, then dy/dx = (1^2)(2*2+1)= 5
at (1,2) sketch a tiny segment that has a slope approx =5
(hint: depending on how exact you need to be, use a ruler to line up (1,2) with ( 2,7), then draw only a small part of the line.

B) since x^2 is >=0, the slope will be positive if 2y+1> 0, so y> -1/2
on the slope field, that would be all points above y= -1/2, except on the y axis, where the slopes are zero.

C) dy/dx= x^2(2y+1)
1/(2y+1) dy= x^2 dx

(1/2) ln(2y+1) = x^3/3+ C

Ln(2y+1)= (2/3)x^3 +C

2y+1= e^[ (2/3)x^3+ C]

= e^C * e^[(2/3)x^3

= Ce^[(2/3)x^3]

Y= {Ce^[(2/3)x^3] -1}/2

= Ce^[(2/3)x^3] -1/2

( notice that C keeps changing, since it is still an unknown constant. e^C or C/2 would be unnecessary)

Now plug in (0,5) to find C

5= Ce^0 -1/2

11/2=C

Y = (11/2) e^ [(2/3)x^3] -1/2

Hoping this helps!
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