Question

equation with the initial condition ( f(1) = 0 ). for this particular solution, ( f(x) lt 1 ) for all values of ( x ).
(a) use eulers method, starting at ( x = 1 ) with two steps of equal size, to approximate ( f(0) ). show the work that leads to your answer.
(b) find ( lim_{x
ightarrow1}\frac{f(x)}{x^{3}-1} ). show the work that leads to your answer.
(c) find the particular solution ( y = f(x) ) to the differential equation ( \frac{dy}{dx}=1 - y ) with the initial condition ( f(1) = 0 ).

Explanation:

Part (a)

Step1: Determine step size

We start at \( x = 1 \) and want to approximate \( f(0) \) with two steps of equal size. The interval is from \( x = 1 \) to \( x = 0 \), so the total change in \( x \) is \( \Delta x_{\text{total}}=0 - 1=- 1 \). For two steps, the step size \( h=\frac{\Delta x_{\text{total}}}{2}=\frac{-1}{2}=-0.5 \)

Step2: First Euler step (from \( x = 1 \) to \( x=1 + h=0.5 \))

We know that the differential equation is \( \frac{dy}{dx}=1 - y \), and the initial condition is \( f(1) = 0 \). Euler's method formula is \( y_{n + 1}=y_{n}+h\times(1 - y_{n}) \)
At \( n = 0 \), \( x_0 = 1 \), \( y_0=f(1) = 0 \)
\( y_1=y_0+h\times(1 - y_0)=0+(-0.5)\times(1 - 0)=- 0.5 \)
So \( f(0.5)\approx - 0.5 \)

Step3: Second Euler step (from \( x = 0.5 \) to \( x=0.5+h = 0 \))

Now, \( n = 1 \), \( x_1 = 0.5 \), \( y_1=-0.5 \)
\( y_2=y_1+h\times(1 - y_1)=-0.5+(-0.5)\times(1-(-0.5))=-0.5+(-0.5)\times1.5=-0.5 - 0.75=-1.25 \)
But wait, the problem states that \( f(x)<1 \) for all \( x \), but our approximation for \( f(0) \) is \( - 1.25 \) (this is still less than \( 1 \)). However, let's check the calculations again. Wait, the differential equation is \( \frac{dy}{dx}=1 - y \), and we are moving from \( x = 1 \) to \( x=0 \), so the step size is negative. The formula for Euler's method when moving in the negative \( x \)-direction is still \( y_{n+1}=y_n+h\times f'(x_n,y_n) \), where \( f'(x,y)=1 - y \)

Wait, maybe I made a mistake in the sign of the step size interpretation. Let's re - express the interval as from \( x = 1 \) to \( x = 0 \), so we can think of it as moving from \( x = 1 \) to \( x=1-0.5 = 0.5 \) (first step) and then from \( x = 0.5 \) to \( x=0.5 - 0.5=0 \) (second step). The formula \( y_{n + 1}=y_n+h\times(1 - y_n) \), where \( h=-0.5 \)

First step: \( x_0 = 1 \), \( y_0 = 0 \)
\( y_1=0+(-0.5)\times(1 - 0)=-0.5 \) (this is \( f(0.5)\approx - 0.5 \))

Second step: \( x_1 = 0.5 \), \( y_1=-0.5 \)
\( y_2=-0.5+(-0.5)\times(1-(-0.5))=-0.5+(-0.5)\times1.5=-0.5-0.75 = - 1.25 \)
So the approximation for \( f(0) \) using two - step Euler's method is \( - 1.25 \)

Part (b)

We know that \( f(1) = 0 \), so we have a limit of the form \( \lim_{x
ightarrow1}\frac{f(x)}{x^{3}-1} \), which is a \( \frac{0}{0} \) indeterminate form (since as \( x
ightarrow1 \), \( x^{3}-1=(x - 1)(x^{2}+x + 1)
ightarrow0 \) and \( f(1) = 0 \)). We can apply L'Hopital's Rule.

Step1: Apply L'Hopital's Rule

By L'Hopital's Rule, if \( \lim_{x
ightarrow a}\frac{g(x)}{h(x)} \) is a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) indeterminate form, then \( \lim_{x
ightarrow a}\frac{g(x)}{h(x)}=\lim_{x
ightarrow a}\frac{g'(x)}{h'(x)} \)
Let \( g(x)=f(x) \) and \( h(x)=x^{3}-1 \)
We know that \( g'(x)=f'(x)=1 - f(x) \) (from the differential equation \( \frac{dy}{dx}=1 - y \)) and \( h'(x)=3x^{2} \)

Step2: Evaluate the new limit

Now we find \( \lim_{x
ightarrow1}\frac{f'(x)}{3x^{2}}=\lim_{x
ightarrow1}\frac{1 - f(x)}{3x^{2}} \)
We can substitute \( x = 1 \) into the function \( \frac{1 - f(x)}{3x^{2}} \) because \( f(1) = 0 \) and the function is continuous (since \( f(x) \) is a solution to a first - order linear differential equation, it is differentiable and hence continuous)
\( \frac{1 - f(1)}{3\times1^{2}}=\frac{1-0}{3}=\frac{1}{3} \)

Part (c)

We have the first - order linear differential equation \( \frac{dy}{dx}+y = 1 \) with the initial condition \( y(1)=0 \) (where \( y = f(x) \))

Step1: Find the integrating factor

The standard form of a first - order linear differential equation is \( \frac{dy}{dx}+P(x)y=Q(x) \). Here, \( P(x) = 1 \) and \( Q(x)=1 \)
The integrating factor \( \mu(x)=e^{\int P(x)dx}=e^{\int 1dx}=e^{x} \)

Step2: Multiply the differential equation by the integrating factor

Multiply \( \frac{dy}{dx}+y = 1 \) by \( e^{x} \):
\( e^{x}\frac{dy}{dx}+e^{x}y=e^{x} \)
The left - hand side is the derivative of \( (e^{x}y) \) with respect to \( x \) by the product rule (since \( \frac{d}{dx}(e^{x}y)=e^{x}y+e^{x}\frac{dy}{dx} \))
So we have \( \frac{d}{dx}(e^{x}y)=e^{x} \)

Step3: Integrate both sides

Integrate both sides with respect to \( x \):
\( \int\frac{d}{dx}(e^{x}y)dx=\int e^{x}dx \)
\( e^{x}y=e^{x}+C \)

Step4: Solve for \( y \)

Divide both sides by \( e^{x} \):
\( y = 1+Ce^{-x} \)

Step5: Apply the initial condition

We know that \( y(1)=f(1) = 0 \). Substitute \( x = 1 \) and \( y = 0 \) into \( y = 1+Ce^{-x} \)
\( 0=1+Ce^{-1} \)
\( Ce^{-1}=-1 \)
\( C=-e \)

Step6: Write the particular solution

Substitute \( C=-e \) into \( y = 1+Ce^{-x} \)
\( y = 1 - e\times e^{-x}=1 - e^{1 - x} \)
So \( f(x)=1 - e^{1 - x} \)

Answer:

s

(a) The approximation of \( f(0) \) using Euler's method with two steps is \( \boldsymbol{-1.25} \)

(b) The value of \( \lim_{x
ightarrow1}\frac{f(x)}{x^{3}-1} \) is \( \boldsymbol{\frac{1}{3}} \)

(c) The particular solution \( f(x) \) is \( \boldsymbol{f(x)=1 - e^{1 - x}} \)