Find the limit?

Question

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.

Answer 1

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:

lim_x→0(y+1)/sin(x) = lim_x→0 y'/cos(x) = lim y²(2x+2)/cos(x) = 2.

 

b) Euler's method tells us that y_n+1=y_n + h y'(x_n,y_n).

If you go in two steps of equal size from 0 to 1/2, the step size h must equal 1/4:

 

y_n+1= y_n + (1/4) y'(x_n,y_n)

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.