DIFFERENTIAL EQUATION 6
DifferentialEquation
DifferentialEquation
3.

y’= y^{2}te^{t}
y=1/2y^{2}t^{2}e^{t }+c
Makingy the subject, y= 2t^{2}e^{t}+ c is a oneparameter family of solutions to y’=y^{2}te^{t}andc is a constant

y’= y^{2}te^{t}
=(2t^{2}e^{t}+ c)^{2}te^{t}= (4t^{2}e^{2t})te^{t}
^{ }But^{}4t^{2}e^{2t}isa square of y. Hence y’=y^{2}te^{t}
c)No solution misses from the set of the oneparameter formula ofsolutions.
d)y(0) = 1
y(t) = 2t^{2}e^{t}+c
^{ }y(0)=2^{(0)}2e^{(0)}+c= 1. There c= 1
Thesolution that satisfies the initial condition y (0)=1 is y= 2t^{2}e^{t}+ 1
e) 2t^{2}e^{t}+ 1= 1
2t^{2}e^{t=}0t^{2}e^{t}=0.Therefore t= {0, ∞}
4.b) y= 2t^{2}e^{t}+ 1. The starting point is t_{0}=0 and y_{0}=1
t_{1}=t_{0}+h= 0+0.5= 0.5
y_{1}=y_{o}+hf (t_{0},y_{0})=1 + 0.5 (2 *0+1)= 1.5
t_{2}=t_{1}+ h= 0.5 + 0.5= 1
y_{2}=y_{1} + hf (t_{1}, y_{1})= 1.5 +0.5(2*0.5+1.5)= 1.7500 The value of y_{1 }at t=1 is 1.7500
h 
t 
Y 
0.5 
0.0000 
1.0000 
1.0 
0.5000 
1.5000 
1.5 
1.0000 
1.7500 

Using the two step Euler method, h=0.5
Step1: y(t_{1})y(t_{0})/t_{1}t_{0}=y’(t_{0})
y(0)=1
y’(0.5)=y(0.5)1/0.5=0 thus
y(0.5)=1
Step2
y(1)y(0.5)/10.5=y’(0.5)
y’(0.5)=(1)^{2}*0.5*e^{0.5}=0.8243
Thereforey (1)= 1.41215
y(1) = 0.45495
h 
t 
y 
0.5 
0.5 
1 
1.0 
1 
1.41215 

There are no issues at t=1 because the curve is tangent to the initial condition of y(0)=1
5a) It is also reasonable to assume that the total time required forthe damaged area to heal depends on the area injured. Thus, anincrease in the time taken to recover implies a larger area wasdamaged.

The boundary of A is measured in A and the time taken to heal is given in seconds or minutes.

The model for size A (t) of the damaged area assumes the differential equation because it represents an exponential curve, where the rate of change in variable A has to be proportional to the variable itself.

A’= kA^{1/2. }Therefore, A= kA^{1/2 }t
Thevalue of A(0) is the area of the damaged that is circular.3.142*4*4 = 50.272cm^{2}
TheDE= 0.0001*50.272^{1/2}=0.00070903 cm^{2}/sec.It implies that the rate at which the damaged area decreases in asecond is 0.00070903 cm^{2.}
A(t) = kA^{1/2}t
Thewound takes 19.7 days to heal.

The answer may not be a good approximate because, despite the shape of the wound, it is difficult to tell how far beneath the upper skin the injury had extended.