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.
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!
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!