Differential Equation

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DIFFERENTIAL EQUATION 6

DifferentialEquation

DifferentialEquation

3.

  1. y’= y2te-t

y=-1/2y2t2e-t +c

Makingy the subject, y= -2t2e-t+ c is a one-parameter family of solutions to y’=y2te-tandc is a constant

  1. y’= y2te-t

=(-2t2e-t+ c)2te-t= (-4t2e-2t)te-t

But-4t2e-2tisa square of y. Hence y’=y2te-t

c)No solution misses from the set of the one-parameter formula ofsolutions.

d)y(0) = 1

y(t) = -2t2e-t+c

y(0)=-2(0)2e(-0)+c= 1. There c= 1

Thesolution that satisfies the initial condition y (0)=1 is y= -2t2e-t+ 1

e) -2t2e-t+ 1= 1

-2t2e-t=0t2e-t=0.Therefore t= {0, ∞}

4.b) y= -2t2e-t+ 1. The starting point is t0=0 and y0=1

t1=t0+h= 0+0.5= 0.5

y1=yo+hf (t0,y0)=1 + 0.5 (-2 *0+1)= 1.5

t2=t1+ h= 0.5 + 0.5= 1

y2=y1 + hf (t1, y1)= 1.5 +0.5(-2*0.5+1.5)= 1.7500 The value of y1 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

  1. Using the two step Euler method, h=0.5

Step1: y(t1)-y(t0)/t1-t0=y’(t0)

y(0)=1

y’(-0.5)=y(-0.5)-1/-0.5=0 thus

y(-0.5)=1

Step2

y(-1)-y(0.5)/-1-0.5=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

  1. 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.

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

  2. 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.

  3. A’= -kA1/2. Therefore, A= -kA1/2 t

Thevalue of A(0) is the area of the damaged that is circular.3.142*4*4 = 50.272cm2

TheDE= -0.0001*50.2721/2=-0.00070903 cm2/sec.It implies that the rate at which the damaged area decreases in asecond is 0.00070903 cm2.

A(t) = -kA1/2t

Thewound takes 19.7 days to heal.

  1. 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.