Question

Find the solution to the initial value problem. $$ x y^{\prime}=y(x-2), y(1)=3 $$

Short answer

The solution is \( y = \frac{3}{x^2} e^{x-1} \).

Step-by-step solution

Simplify the Differential Equation

Rewrite the given differential equation in a more standard form. Start with the equation:\[ x y' = y(x-2) \]Divide both sides by \( y \) (assuming \( y eq 0 \)) to obtain:\[ \frac{y'}{y} = \frac{x-2}{x} \]

Separate Variables

To solve the differential equation by separation of variables, isolate terms involving \( y \) on one side and \( x \) on the other:\[ \frac{y'}{y} = \frac{x-2}{x} \]

Integrate Both Sides

Integrate both sides of the equation:\[ \int \frac{dy}{y} = \int \left( 1 - \frac{2}{x} \right) \, dx \]This gives:\[ \ln|y| = x - 2\ln|x| + C \]

Solve for y

Exponentiate both sides to solve for \( y \):\[ |y| = e^{x - 2\ln|x| + C} \]This can be rewritten as:\[ |y| = C'e^x \frac{1}{x^2} \]where \( C' = e^C \) is a constant.

Apply the Initial Condition

Use the initial condition \( y(1) = 3 \) to solve for the constant \( C' \):Substitute \( x = 1 \) and \( y = 3 \) into the equation:\[ 3 = C' \cdot \frac{e^1}{1^2} \]which simplifies to:\[ 3 = C'e \] Thus,\[ C' = \frac{3}{e} \].

Write the Final Solution

Substitute \( C' = \frac{3}{e} \) back into the equation:\[ y = \frac{3}{e} e^x \frac{1}{x^2} = \frac{3}{x^2} e^{x-1} \]This is the particular solution that satisfies the initial condition.

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation where, in addition to the equation itself, we are also given a condition that specifies the value of the unknown function at a particular point. In our exercise, the differential equation is connected with an initial condition, which is \( y(1) = 3 \). This means that when \( x = 1 \), the function \( y \) equals 3.
This condition is crucial as it allows us to determine the specific solution to the problem among a family of potential solutions. Initial conditions help narrow down the infinite possibilities to a singular solution that satisfies both the differential equation and the initial condition.

Separation of Variables

Separation of variables is a fundamental method used to solve first-order differential equations where the goal is to rearrange the equation so that each variable is on a separate side of the equation. In the given exercise, the equation \( xy' = y(x-2) \) was rearranged into the form \( \frac{y'}{y} = \frac{x-2}{x} \).
By applying separation of variables, the problem becomes much more approachable because it allows us to integrate with respect to \( y \) on one side and \( x \) on the other. This technique is particularly useful for equations that can be manipulated into a product of functions of only one variable each.

Integration Techniques

Integration techniques are essential tools in solving differential equations, especially when using the method of separation of variables. In the exercise, we needed to integrate both sides of the equation: \( \int \frac{dy}{y} = \int \left( 1 - \frac{2}{x} \right) \, dx \).

• For the left side, \( \int \frac{dy}{y} \), the integration results in \( \ln|y| \), familiar from the natural logarithm integration rule.
• For the right side, the integration involves splitting into two terms: \( \int 1 \, dx \) which gives \( x \), and \( \int \frac{-2}{x} \, dx \) which results in \( -2\ln|x| \).

Combining these results enables us to build the expression \( \ln|y| = x - 2\ln|x| + C \), where \( C \) is the constant of integration.

Exponential Function

To isolate \( y \) in the solution process, we need to undo the natural logarithm by applying the exponential function. This transforms \( \ln|y| = x - 2\ln|x| + C \) into \( |y| = e^{x - 2\ln|x| + C} \).
The exponential function thus converts our logarithmic expression into a multiplicative one, allowing us to express \( |y| \) in terms of \( e^x \) and power functions of \( x \). This is rewritten as \( |y| = C'e^x \frac{1}{x^2} \) by recognizing that \( C' = e^C \).
This is a common technique, leveraging the properties of natural logarithms and exponentials to simplify and solve equations.

Constant of Integration

The constant of integration, denoted as \( C \) in indefinite integrals, plays a critical role in finding the particular solution to differential equations. In our exercise, the integration resulted in \( \ln|y| = x - 2\ln|x| + C \), but it's \( C' = e^C \) that fits into the exponential solution.
Applying the initial condition \( y(1) = 3 \) is used to find the value of \( C' \). Substituting these values into the equation, we compute \( C' = \frac{3}{e} \) by solving \( 3 = C'e \).
This calculated \( C' \) is then incorporated into the final expression for \( y \), ensuring the solution satisfies the initial value condition entirely, yielding \( y = \frac{3}{x^2} e^{x-1} \).
Understanding the role of the constant of integration is crucial since it often serves as the bridge from the general solution to a particular solution that meets specified criteria.