Problem 30 - Advanced Engineering Mathematics

Question

Separating variables, we have \\[\frac{d y}{y^{2}-y}=\frac{d x}{x} \quad \text { or } \quad \int \frac{d y}{y(y-1)}=\ln |x|+c.\\] Using partial fractions, we obtain $$\begin{aligned} \int\left(\frac{1}{y-1}-\frac{1}{y}\right) d y &=\ln |x|+c \\ \ln |y-1|-\ln |y| &=\ln |x|+c \\ \ln \left|\frac{y-1}{x y}\right| &=c \\ \frac{y-1}{x y} &=e^{c}=c_{1}.\end{aligned}$$ Solving for \(y\) we get \(y=1 /\left(1-c_{1} x\right) .\) We note by inspection that \(y=0\) is a singular solution of the differential equation. (a) Setting \(x=0\) and \(y=1\) we have \(1=1 /(1-0),\) which is true for all values of \(c_{1} .\) Thus, solutions passing through (0,1) are \(y=1 /\left(1-c_{1} x\right)\). (b) Setting \(x=0\) and \(y=0\) in \(y=1 /\left(1-c_{1} x\right)\) we get \(0=1 .\) Thus, the only solution passing through (0,0) is \(y=0\). (c) Setting \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\) we have \(\frac{1}{2}=1 /\left(1-\frac{1}{2} c_{1}\right),\) so \(c_{1}=-2\) and \(y=1 /(1+2 x)\). (d) \(\operatorname{Setting} x=2\) and \(y=\frac{1}{4}\) we have \(\frac{1}{4}=1 /\left(1-2 c_{1}\right),\) so \(c_{1}=-\frac{3}{2}\) and \(y=1 /\left(1+\frac{3}{2} x\right)=2 /(2+3 x)\).

Step-by-Step Solution

Answer

The solutions are \( y = \frac{1}{1 - c_1 x} \) and \( y = 0 \), with specific cases: \( c_1 = -2 \) for \( (\frac{1}{2}, \frac{1}{2}) \) and \( c_1 = -\frac{3}{2} \) for \( (2, \frac{1}{4}) \).

Step 1: Separation of Variables

Starting with the given differential equation \( \frac{d y}{y^2 - y} = \frac{d x}{x} \), separate the variables by rearranging terms to get \( \int \frac{d y}{y(y-1)} = \ln |x| + c \). This sets up the equation for integrating each variable independently.

Step 2: Partial Fraction Decomposition

Decompose the integrand \( \frac{1}{y(y-1)} \) into partial fractions: \( \frac{1}{y} - \frac{1}{y-1} \). Now, rewrite the equation: \( \int \left( \frac{1}{y-1} - \frac{1}{y} \right) d y = \ln |x| + c \).

Step 3: Integrate Both Sides

Integrate both sides: \( \ln |y-1| - \ln |y| = \ln |x| + c \). Use properties of logarithms to simplify: \( \ln \left| \frac{y-1}{y} \right| = \ln |x| + c \).

Step 4: Exponentiation

Exponentiate both sides to eliminate the natural logarithm: \( \frac{y-1}{y} = c_1 x \), where \( e^c = c_1 \) is a constant. Rearrange to solve for \( y \): \( y = \frac{1}{1 - c_1 x} \).

Step 5: Consider Singular Solutions

Inspect for any additional singular solutions. By inspection, \( y = 0 \) is a singular solution of the original differential equation \( \frac{d y}{y^2 - y} = \frac{d x}{x} \).

Step 6: Verify Specific Solutions

(a) For initial condition \( (0, 1) \), \( 1 = \frac{1}{1 - 0} \) holds, so any \( c_1 \) is valid. The general solution becomes \( y = \frac{1}{1 - c_1 x} \). (b) For initial condition \( (0, 0) \), the only solution is \( y = 0 \). (c) For \( (\frac{1}{2}, \frac{1}{2}) \), solve \( \frac{1}{2} = \frac{1}{1 - \frac{1}{2} c_1} \) resulting in \( c_1 = -2 \), and \( y = \frac{1}{1 + 2x} \). (d) For \( (2, \frac{1}{4}) \), solve \( \frac{1}{4} = \frac{1}{1 - 2c_1} \) yielding \( c_1 = -\frac{3}{2} \), leading to \( y = \frac{2}{2 + 3x} \).

Key Concepts

Separation of Variables

Separation of variables is a fundamental technique used to solve differential equations. It involves rearranging the equation such that each variable and its differential are on different sides of the equation. This method is particularly effective when dealing with first-order ordinary differential equations (ODEs).

In our example, we start with the equation \( \frac{d y}{y^2 - y} = \frac{d x}{x} \). The goal is to express the equation such that all terms involving \( y \) are on one side and all terms involving \( x \) are on the other. This rearrangement allows us to integrate each side independently, leading to solutions that describe how \( y \) changes with \( x \).

This technique is powerful because it transforms a differential equation into simpler integrals that can often be solved using basic calculus techniques. Once variables are separated, we can integrate each side to find a general solution, often involving arbitrary constants that can be further determined using initial conditions or boundary values.

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic technique used to break down complex rational expressions into simpler parts. This method is especially useful when integrating rational functions, as it simplifies the integrals into more manageable pieces.

In the context of solving our differential equation, after separating variables, we encounter the integrand \( \frac{1}{y(y-1)} \). To integrate this, we apply partial fraction decomposition:

• Recognize that \( \frac{1}{y(y-1)} \) can be decomposed into \( \frac{A}{y} + \frac{B}{y-1} \).
• Solve for \( A \) and \( B \) by clearing the fractions and equating coefficients.
• In our problem, it simplifies to \( \frac{1}{y-1} - \frac{1}{y} \).

This decomposition lets us integrate each term separately, significantly simplifying the process. Partial fractions are a handy tool for making integration more approachable, especially with complex expressions.

Singular Solutions

Singular solutions are particular solutions to a differential equation that do not fit the form of the general solution. They are important because they represent unique cases that might not be obvious from the general solution alone.

In our problem, we have \( \frac{d y}{y^2 - y} = \frac{d x}{x} \), leading to the general solution: \( y = \frac{1}{1 - c_1 x} \). However, upon inspection, we notice that \( y = 0 \) also satisfies the differential equation. This solution does not fit the general form derived from our method but is still valid.

Singular solutions might arise from specific initial or boundary conditions and are crucial in ensuring that all possible solutions to a differential equation are accounted for. They provide deeper insight into the behavior of a system described by the equation and are vital for a complete understanding of the solution set.