Question

consider the differential equation \\( \frac{d y}{d x}=1 - y \\). let \\( y = f(x) \\) be the particular solution to this differential equation with the initial condition \\( f(1)=0 \\). for this particular solution, \\( f(x)<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 \to 1} \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{d y}{d x}=1 - y \\) with the initial condition \\( f(1)=0 \\).

Explanation:

Part (a)

Step1: Determine step size

We want to go from \( x = 1 \) to \( x = 0 \) in two steps of equal size. The total interval is \( 0 - 1=- 1 \), so each step size \( h=\frac{-1}{2}=-0.5 \)

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

We know that \( y_0=f(1) = 0 \), and the differential equation is \( \frac{dy}{dx}=1 - y \). At \( x = 1,y = 0 \), the slope \( m_0=1 - y_0=1-0 = 1 \)

Using Euler's formula \( y_{n+1}=y_n+h\times m_n \), for \( n = 0 \):
\( y_1=y_0+h\times m_0=0+(-0.5)\times1=- 0.5 \)

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

At \( x = 0.5,y = y_1=-0.5 \), the slope \( m_1=1 - y_1=1-(-0.5)=1.5 \)

Using Euler's formula \( y_{2}=y_1+h\times m_1=-0.5+(-0.5)\times1.5=-0.5 - 0.75=-1.25 \)

So the approximation of \( f(0) \) is \( y_2=-1.25 \)

Step1: Analyze the limit form

We want to find \( \lim_{x
ightarrow1}\frac{f(x)}{x^3 - 1} \). First, check the form of the limit as \( x
ightarrow1 \):

When \( x
ightarrow1 \), \( f(1) = 0 \) and \( x^3-1=(x - 1)(x^2+x + 1)
ightarrow0 \), so we have a \( \frac{0}{0} \) indeterminate form, and we can use L'Hopital's Rule.

Step2: Apply L'Hopital's Rule

By L'Hopital's Rule, \( \lim_{x
ightarrow1}\frac{f(x)}{x^3 - 1}=\lim_{x
ightarrow1}\frac{f^{\prime}(x)}{3x^2} \)

Since \( y = f(x) \) satisfies \( \frac{dy}{dx}=1 - y \), so \( f^{\prime}(x)=1 - f(x) \)

Step3: Substitute \( x = 1 \)

As \( x
ightarrow1 \), \( f(x)
ightarrow f(1) = 0 \), so \( f^{\prime}(1)=1 - f(1)=1-0 = 1 \)

Then \( \lim_{x
ightarrow1}\frac{f^{\prime}(x)}{3x^2}=\frac{f^{\prime}(1)}{3\times1^2}=\frac{1}{3} \)

Step1: Solve the differential equation

The differential equation is \( \frac{dy}{dx}=1 - y \), which is a separable differential equation.

Separate the variables: \( \frac{dy}{1 - y}=dx \)

Step2: Integrate both sides

Integrate left - hand side with respect to \( y \) and right - hand side with respect to \( x \):

\( \int\frac{dy}{1 - y}=\int dx \)

Let \( u = 1 - y \), then \( du=-dy \), so \( \int\frac{dy}{1 - y}=-\int\frac{du}{u}=-\ln|u|+C=-\ln|1 - y|+C \)

\( \int dx=x + C \)

So we have \( -\ln|1 - y|=x + C \)

Multiply both sides by \(- 1\): \( \ln|1 - y|=-x + C_1 \) (where \( C_1=-C \))

Exponentiate both sides: \( |1 - y|=e^{-x + C_1}=e^{C_1}e^{-x} \)

We can write \( 1 - y=Ke^{-x} \) (where \( K=\pm e^{C_1} \), and we can absorb the sign into \( K \))

Then \( y = 1 - Ke^{-x} \)

Step3: Apply the initial condition

Use the initial condition \( f(1)=0 \), substitute \( x = 1,y = 0 \) into \( y = 1 - Ke^{-x} \):

\( 0=1 - Ke^{-1} \)

\( Ke^{-1}=1 \)

\( K = e \)

So the particular solution is \( y = 1 - e\times e^{-x}=1 - e^{1 - x} \)

Answer:

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

Part (b)