Consider the differential equation for the function y given by y\' = -5 y^2. a.
ID: 2893747 • Letter: C
Question
Consider the differential equation for the function y given by y' = -5 y^2. a. Find the implicit expression of all solutions y of the equation above in the form psi(t, y) = c, with the normalization psi(0, 1) = -1/1, psi(t, y) = b. Find the explicit expression for a solution y_1 found in (a) satisfying the initial condition y_1(0) = 1, y_1(t) = c. Find the value t_1 where the solution y_1 is not defined, t_1 = d. Find the explicit expression for a solution y_2 found in (a) satisfying the initial condition y_2(0) = 2, y_2(t) = e. Find the value t_2 where the solution y_2 is not defined, t_2 = We have solved a differential equation with two different initial conditions. The two solutions obtained are defined on different domains. This is an example of solutions to non-linear differential equations with domain depending on their initial data.Explanation / Answer
given y'=-5y2
dy/dt =-5y2
seperate the variables
dy/y2 =-5dt
integrate on both sides
dy/y2 =-5dt
=>(-1/y)=-5t +c
=>(1/y)=5t -c
=>(1/y)-5t +c=0
(t,y)=(1/y)-5t +c
a)
(0,1) =-1/1
=>(1/1)-5*0 +c=-1/1
=>1-0 +c=-1
=>c=-2
(t,y)=(1/y)-5t -2
(b)
(1/y)=5t -c
=>y=1/(5t-c)
y(0)=1
=>1=1/(5*0 -c)
=>1=-1/c
=>c=-1
=>y=1/(5t+1)
so y1(t)=1/(5t+1)
(c)
y1(t) is not defined when 5t +1=0
=>t=-1/5
(d)
y(0)=2
=>2=1/(5*0 -c)
=>c=-1/2
=>y=1/(5t-(-1/2))
=>y=2/(10t+1)
y2(t) =2/(10t+1)
(e)
y2(t) is not defined when 10t +1=0
=>t=-1/10