A function y(t) satisfies the diff eq: dy/dt= (-y^4)+(5y^3)+(36y^2)
a.) what are the constant solutions of this equation?
b.) for what values of y is y increasing?
please help!
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To find the steady state solutions (i.e. y = constant) we look for where dy/dx=0. Factoring the right hand side:
dy/dx = -y^2(y-9)(4+4)
so constant solutions are y=0, y=9, y=-4
To determine where y would be increasing, we determine where dy/dx > 0.
Using the zeroes determined above, we find that
dy/dx > 0 if -4 < y < 0 or 0 < y < 9
So y would be increasing if y is on (-4,0) U (0, 9).
Update*******
Check the answer. If y < -4, dy/dx is negative.
a. To get the constant solutions, simply equate dy/dt to 0. This is obvious because the derivative of a constant is 0. So dy/dt=-y^2(y^2-5y-36)=0 –> y=-4, 0, 9 for all t element of real numbers
b. y is increasing if the first derivative is greater than 0. Thus, dy/dt=-y^2(y^2-5y-36)>0 –>y<-4 or 0
a) dy/dt = -y^4 + 5y^3 + 36y^2
dy/(-y^4 + 5y^3 + 36y^2) = dt
dy/[-y^2(y^2 - 5y - 36)] = dt
dy/[-y^2(y - 9)(y + 4)] = dt
1/[-y^2(y - 9)(y + 4)] = A/y + B/y^2 + C/(y - 9) + D/(y + 4)
-1 = Ay(y - 9)(y + 4) + B(y - 9)(y + 4) + Cy^2(y + 4) + Dy^2(y - 9)
y = 0: -1 = -36B, so B = 1/36
y = -4: -1 = -208D, so D = 1/208
y = 9: -1 = 1053C, so C = -1/1053
y = 1: -1 = -40A + (-32)(1/36) + 5(-1/1053) + (-8)(1/208)
-1 = -40A - 8/9 - 5/1053 - 1/26
-11/162 = -40A
A = -11/6480
So, -11/6480 ∫ dy/y + 1/36 ∫ dy/y^2 - 1/1053 ∫ dy/(y - 9) + 1/208 ∫ dy/(y + 4) =
-11/6480 ln y - 1/(36y) - 1/1053 ln (y - 9) + 1/208 ln (y + 4) = t + C
This is absolutely monstrous! I can’t believe this is even close to the answer. Maybe this will give you an idea to work on.
This is a separable equation, and can be put in the form
dy/[y^2(9 - y)(4 + y)] = dt. Use a partial fraction decomposition to find
1/[y^2(9 - y)(4 + y)] = A/y + B/y^2 + C/(9 - y) + D/(4 + y). Integration will lead to
Aln|y| - B/y - Cln|9 - y| + Dln|4 + y| = t + K. You can fill in the rest.
dy/dt= 0; hence, -y^4+5y^3+36^2= 0
5y^3+36^2=y^4
divide throughout by y^2
5y+36=y^2
re-arrange
y^2-5y-36=0, using almighty formular i.e [-b+/-(b^2-4ac)^1/2]/2a
a=1, b=-5 and c=-36
y=-4 or 6
b. y= 6