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: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 [tex]\frac{dy}{dt} =y^4-6y^2+5y^2\\=y^2(y^2-6y+5)\\=y^2(y-1)(y-5)[/tex]Separate the variables to have[tex]\frac{dy}{y^2(y-1)(y-5)} =dt[/tex]Left side we can resolve into partial fractionsLet [tex]\frac{1}{y^2(y-1)(y-5)} =\frac{A}{y} +\frac{B}{y^2}+\frac{C}{y-1} \frac{D}{y-5}[/tex]Taking LCD we get[tex]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\\[/tex]By equating coeff of y^3 we haveA+C+D=0[tex]C=\frac{-1}{4} \\D=\frac{1}{100} \\B =\frac{1}{5} \\A = -C-D = \frac{6}{25}[/tex]Hence left side =[tex]\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[/tex]b) y is increasing whenever dy/dt>0dy/dt =0 at points y =0, 1 and 5dy/dt >0 in (-\infty,0) U (1,5)Hence increasing in  (-\infty,0) U (1,5)c) Decreasing in (0,1)