Question

A function y(t) satisfies the differential equation dy dt = y 4 − 6y 3 + 5y 2 . (a) What are the constant solutions of the equation? (Recall that these have the form y = C for some constant, C.) (b) For what values of y is y increasing? (c) For what values of y is y decreasing?

Answer 1

Answer:

a) y = 0 , 5,1

b) y ⊂ (- ∞,0) ∪ (0,1)∪(5,∞)

Step-by-step explanation:

Given data:

differential equation is given as

$\frac{dy}[dt} = y^4 -6y^3+ 5y^2$

a) constant solution

$y^4 -6y^3+ 5y^2 = 0$

taking y^2 from all part

$y^2(y^2 - 6y -5) = 0$

solution of above equation is

y = 0 , 5,1

b) for which value y is increasing

$\frac{dy}{dt} > 0$

y^2(y - 5) (y -1) > 0

y ⊂ (- ∞,0) ∪ (0,1)∪(5,∞)

Answer 2

Answer:

Hence increasing in  (-\infty,0) U (1,5)

c) Decreasing in (0,1)

Step-by-step explanation:

Given that y(t) satisfies the differential equation

$\frac{dy}{dt} =y^4-6y^2+5y^2\\=y^2(y^2-6y+5)\\=y^2(y-1)(y-5)$

Separate the variables to have

$\frac{dy}{y^2(y-1)(y-5)} =dt$

Left side we can resolve into partial fractions

Let $\frac{1}{y^2(y-1)(y-5)} =\frac{A}{y} +\frac{B}{y^2}+\frac{C}{y-1} \frac{D}{y-5}$

Taking LCD we get

$1= Ay(y-1)(Y-5) +B(y-1)(y-5)+Cy^2 (y-5)+Dy^2 (y-1)\\Put y =1\\1 = -4C\\Put y =5\\ 1 = 25(4)D\\Put y =0\\1=5B$

By equating coeff of y^3 we have

A+C+D=0

$C=\frac{-1}{4} \\D=\frac{1}{100} \\B =\frac{1}{5} \\A = -C-D = \frac{6}{25}$

Hence left side =

$\frac{6}{25y} +\frac{1}{5y^2}+\frac{-1}{4(y-1)}+ \frac{1}{100(y-5)}=dt\\\frac{6}{25}ln y -\frac{1}{5y}-\frac{1}{4}ln|(y-1)| +\frac{1}{100}ln|y-5| = t+C$

b) y is increasing whenever dy/dt>0

dy/dt =0 at points y =0, 1 and 5

dy/dt >0 in (-\infty,0) U (1,5)

Hence increasing in  (-\infty,0) U (1,5)

c) Decreasing in (0,1)