Question

Find the solution to the initial value problem. $$ (x-1) y^{\prime}=y-2, y(0)=0 $$

Short answer

The solution is \(y = -2|x-1| + 2\).

Step-by-step solution

Rewrite the Differential Equation

The given differential equation is \((x-1) y' = y - 2\). First, we rewrite it in standard form by expressing it as a separable differential equation. Begin by dividing both sides by \(x-1\), as long as \(x eq 1\), to isolate \(y'\) on one side:\[y' = \frac{y-2}{x-1}.\]

Separate Variables

Our goal is to separate the variables \(x\) and \(y\). This involves rearranging the equation such that all terms involving \(y\) are on one side and those involving \(x\) are on the other:\[\frac{y-2}{y'} = x-1.\]Separating variables gives:\[\frac{1}{y-2} dy = \frac{1}{x-1} dx.\]

Integrate Both Sides

Integrate both sides of the equation:\[\int \frac{1}{y-2} \, dy = \int \frac{1}{x-1} \, dx.\]The integrals become:\[\ln |y-2| = \ln |x-1| + C,\]where \(C\) is the constant of integration.

Solve for y

Exponentiate both sides to solve for \(y\):\[|y-2| = e^C |x-1|.\]Let \(e^C = C_1 > 0\), then\[y = C_1 |x-1| + 2.\]

Apply Initial Condition

Use the initial condition \(y(0) = 0\) to find \(C_1\). Substitute \(x = 0\) and \(y = 0\) into the solution:\[0 = C_1 |0-1| + 2.\]This simplifies to:\[0 = C_1 \cdot 1 + 2,\]which gives \(C_1 = -2\). Therefore, the specific solution is:\[y = -2|x-1| + 2.\]

Verify the Solution

To verify, substitute \(y = -2|x-1| + 2\) back into the original differential equation and the initial condition. Note that \(y' = 2 \) for \(x<1\) or \(x>1\). For \(y(0) = 0\), verify:\[\text{Substitute } x=0: \quad y=-2(1)+2 = 0,\]which satisfies the initial condition.This verifies \(y = -2|x-1| + 2\) as the solution.

Key concepts

Differential Equation

A differential equation is an equation involving a function and its derivatives. It describes how a quantity changes with others by expressing relationships between these quantities.
In the given problem, the differential equation is \((x-1) y' = y - 2\). Here, we have a relationship between the function \(y\) and its derivative \(y'\). This suggests how \(y\), a function of \(x\), changes as \(x\) changes.

• The term \((x-1)\) is multiplied by \(y'\), indicating that the rate of change of \(y\) depends on \(x\).
• The right side of the equation, \(y - 2\), points out a simple difference that the change is adjusting for a fixed value \(2\).

Differential equations are fundamental in modeling real-world phenomena where rates of change are constant areas of interest, such as physics, engineering, and economics.

Separable Variables

The method of separable variables is a technique used to solve a differential equation by separating the variables \(x\) and \(y\) on different sides of the equation. The goal is to make each side depend only on one variable.
In the differential equation \( (x-1) y' = y - 2 \), we separate the variables by initially rewriting it as \( y' = \frac{y-2}{x-1} \). This allows us to express it as:

• \( \frac{1}{y-2} dy = \frac{1}{x-1} dx \)

The core idea is that by separating these variables, we are preparing the equation for integration. Each side of the equation now independently refers to either \(y\) or \(x\), simplifying the process of finding a solution. Separable variables are especially useful when dealing with multiplicative relationships in first-order differential equations.

Integration

Integration is the process used to find the original function from its derivative. It plays a crucial role in solving differential equations as it helps to 'accumulate' the changes described by the derivative.
Once we have separated the variables in our equation, the next step is to integrate both sides:

• Integrate \( \int \frac{1}{y-2} dy \)
• Integrate \( \int \frac{1}{x-1} dx \)

These integrals yield functions in terms of \(ln|y-2|\) and \(ln|x-1|\) respectively, resulting in \( \ln |y-2| = \ln |x-1| + C \), where \(C\) is the constant of integration.
This process of integration transforms the equation into a form that links \(y\) with \(x\), allowing us to solve for the function \(y\). Integration helps us reverse the differentiation process, discovering the behavior of \(y\) over \(x\).

Initial Condition

An initial condition acts like a starting point for solving a differential equation. It gives us the necessary information to find the specific solution from a general solution.
For the problem \( (x-1) y' = y - 2 \) with the initial condition \( y(0) = 0 \), we substitute \(x = 0\) and \(y = 0\) into our general solution.

• The general solution, \(y = C_1 |x-1| + 2\), becomes \(0 = C_1 |0-1| + 2\)
• This calculation helps us find the constant \(C_1\); specifically, \(C_1 = -2\).

The initial condition ensures that the solution satisfies both the differential equation and the starting value. This is crucial for applying the solution to real-world problems where initial states are known.In summary, initial conditions tailor a differential equation’s solution to fit a unique scenario, helping us solve for precisely one specific result.