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Find the limit?

Consider the differential equation dy/dx=y^2(2x+2). Let y=f(x) be the particular solution to the differential equation with initial condition f(0)=-1.
 
a) Find the limit of (f(x)+1)/sin(x) as x approaches to 0. Show the work that leads to your answer.
b) Use Euler's method, starting at x=0 with two steps of equal size, to approximate f(1/2).
c) Find y=f(x), the particular solution to the differential equation with initial condition f(0)=-1.
 
Please show all your work.
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1 Answer

a) The limit of (y+1)/sin(x) as x→0 is of the indeterminate form 0/0, so you need to use l'Hospital's rule:
limx→0(y+1)/sin(x) = limx→0 y'/cos(x) = lim y²(2x+2)/cos(x) = 2.
 
b) Euler's method tells us that yn+1=yn + h y'(xn,yn).
If you go in two steps of equal size from 0 to 1/2, the step size h must equal 1/4:
 
yn+1= yn + (1/4) y'(xn,yn)
y1= y0 + (1/4) y'(x0,y0) = -1+(1/4) 2 = -1/2 ≈ y(1/4)
y2= y1 + (1/4) y'(x1,y1) = -1/2 + (1/4)(5/8) = -11/32  ≈ y(1/2).
 
c) The equation is separable:
 
dy/y² = (2x+2) dx ⇒ -1/y = x²+2x+c ⇒ y = -1/(x²+2x+c)
 
With y(0)=-1 we get c=1, so that
 
y  = -1/(x²+2x+1).
 
Note that y(1/2) = -4/9.

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