1. (Consider the differential equationdy/dt=y(1-y^2)(i) Determine the equilibrium solutions.(ii) Obtain the general solution.(iii) Solve the initial-value problem y(0) = 1/2(iv) Draw the phase line.(v) Sketch solution curves in the ty-plane corresponding to the initial conditions y(0) =-3/2; y(0) =1/2;y(0)=3/22. Consider the differential equationDy/dt= -2ty^2.(i) Obtain the general solution.(ii) Find all values of y0 such that the solution to the initial-value problem y(-1) = y0 isdefined for all real t (i. e., find all y0 such that the solution does not blow up in finite time.)

attached below is the detailed solution

i) equilibrium solution : y = 0 , y^2 = 1 , y = -1 , +1

ii) attached below

iii) attached below

iv) attached below

v) attached below

Step-by-step explanation:

v) For the sketch of curve attached

red curve ( y(0) ) = -3/2

blue curve ( y(0) ) = 1/2

green solution curve ( y(0) ) = 3/2

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

w< 4

step-by-step explanation: