Problem 26 - Calculus: Early Transcendental Functions

Question

Solve the IVP, explicitly if possible. $$y^{\prime}=\frac{x-1}{y}, y(0)=-2$$

Step-by-Step Solution

Answer

The explicit solution to the initial value problem \(y'(x) = (x-1)/y\) with \(y(0) = -2\) is \(y^2/2 = x^2/2 - x +2\).

Step 1: Separate the Variables

Rewrite the differential equation to isolate the variables. Given the equation \(y'(x) = (x-1)/y\), it can be rewritten as \(y dy = (x-1) dx\).

Step 2: Integrate Both Sides

Perform integration on both sides of the equation. After integrating, you'll have \(\int y dy = \int (x-1) dx\), which simplifies to \(y^2/2 = x^2/2 - x + C\). Here, \(C\) is the constant of integration obtained from the process of integrating.

Step 3: Solve for \(C\) Using the Initial Condition

Use the initial condition \(y(0)=-2\) to solve for \(C\). Plugging these values into the equation from Step 2 gives: \((-2)^2/2 = 0^2/2 - 0 + C\), which simplifies to \(C=2\). Then, the solution to our initial value problem becomes \(y^2/2 = x^2/2 - x +2\).

Key Concepts

Separation of Variables

Separation of variables is a powerful technique for solving certain types of differential equations. It involves rearranging the equation so that all terms involving the dependent variable (in this case, \(y\)) are on one side of the equation, and all terms involving the independent variable (\(x\)) are on the other. This allows for easier integration of each variable separately. In this exercise, you began with the differential equation \( y'(x) = \frac{x-1}{y} \). By multiplying both sides by \(y\) and \(dx\), you separated the variables as \( y \, dy = (x-1) \, dx \). This step ensures each variable is isolated, which prepares the equation for the next step: integration. Remember that the key is to maintain equality and ensure that both sides depend only on their respective variables. This separation is crucial for facilitating the solution of differential equations like the one in this initial value problem (IVP).

Integration

Integration is the next step after successfully separating the variables in a differential equation. It involves calculating the antiderivative of each side of the equation with respect to its variable. This can be thought of as 'undoing' differentiation to recover a function from its rate of change. In our example, after separating the variables, we can individually integrate both sides of \( y \, dy = (x-1) \, dx \).

• On the left side, integrating \(y \, dy\) yields \( \frac{y^2}{2} \).
• On the right side, integrating \((x-1) \, dx\) yields \( \frac{x^2}{2} - x \).

With both sides integrated, you end up with \( \frac{y^2}{2} = \frac{x^2}{2} - x + C \), where \(C\) is a constant of integration. This technique is foundational in solving initial value problems, as it transforms a differential equation into a solvable algebraic equation.

Constant of Integration

The constant of integration, denoted as \(C\), plays a vital role whenever you integrate a function. This constant arises because integration is the reverse operation of differentiation, and different functions can have the same derivative. Thus, \(C\) represents the family of all possible solutions.In an initial value problem (IVP), like the one we tackled, the constant is determined by applying the provided initial condition. Here, you used the condition \(y(0) = -2\) to find \(C\). Substitute \(x = 0\) and \(y = -2\) into the integrated equation \( \frac{y^2}{2} = \frac{x^2}{2} - x + C \), resulting in:\[\frac{(-2)^2}{2} = \frac{0^2}{2} - 0 + C \right)\]This simplifies to \(2 = 0 + C\), hence \(C = 2\).
By inserting this specific \(C\) value back into the general solution, you get \( \frac{y^2}{2} = \frac{x^2}{2} - x + 2 \). This process allows the solution to precisely match the initial condition, converting a general solution into a specific one that satisfies the initial constraints of the problem.