Problem 22 - Calculus: Early Transcendental Functions

Question

The differential equation is separable. Find the general solution, in an explicit form if possible. Sketch several members of the family of solutions. $$y^{\prime}=\frac{\left(y^{2}+1\right) \ln x}{4 y}$$

Step-by-Step Solution

Answer

General solution of the differential equation is \(2y^2x^2 - y^2x \ln(x) - y^2x- x \ln(x) + x = C\). Getting an explicit \(y\) value in terms of \(x\) could be quite complex and may not be possible.

Step 1: Separate variables

Rewrite the given differential equation in a form that allows the variables to be separated, i.e., \( y' = \frac{(y^2+1)\ln(x)}{4y}\). Multiply both sides by \(x\), and multiply the \(y\) term to the LHS, which results in \(x \cdot y' = \frac{(y^2+1)\ln(x)}{4}\). Multiply both sides by \(4\) and re-arrange to obtain a term that has only \(y\) on one side and \(x\) on the other: \(4 \cdot y \cdot x \cdot y'= y^2\ln(x) + \ln(x)\).

Step 2: Integrate both sides

On integrating both sides with respect to \(x\), the LHS becomes \(2y^2x^2 = \int y^2\ln(x) dx + \int \ln(x) dx\). Now, the RHS becomes a bit tricky. We have to use integration by parts to solve these integrals.

Step 3: Use integration by parts

Integration by parts is given by \(\int u dv = u \cdot v - \int v du\) where \(u\) and \(dv\) are parts of the integral that we choose. Solving \(\int y^2\ln(x) dx + \int \ln(x) dx\) is done by parts taking \(u = \ln(x)\) and \(dv = dx\). The integral becomes \((y^2x \ln(x) - y^2x) + (x \ln(x) - x) + C\), where \(C\) is the constant of integration.

Step 4: Solve the resulting equation to find \(y\)

Equating both sides, the final result after simplifying is \[2y^2x^2 - y^2x \ln(x) - y^2x- x \ln(x) + x = C\] a solution for \(y\). Now, if it is need to find an explicit solution, that is isolate \(y\), it might be complex to do from this equation, since it's a quartic equation. So, in most situations, you might not be able to get an explicit solution.

Key Concepts

Integrating Differential Equations

Integrating differential equations is a cornerstone concept when solving these mathematical constructs. The process involves finding a function (or functions) that satisfy the relationship expressed by the differential equation. In the case of separable differential equations, the equation can be manipulated into a form where the variables can be separated on either side of the equal sign. This separation allows us to integrate each side with respect to its own variable.

For the given exercise, \( y^\prime = \frac{(y^2 + 1) \ln x}{4 y} \) is separated into terms involving \(y\) and \(x\) before integrating. The integration process involves finding the antiderivative of each side, which yields the general solution. This step requires knowledge of integral calculus and often the use of fundamental integration techniques such as substitution or by parts.

Integration by Parts

Integration by parts is a method used to integrate products of functions and is based on the product rule for differentiation. The formula for integration by parts is \(\int u dv = u v - \int v du\) where \(u\) and \(dv\) are parts of the integral chosen strategically, often with the goal of simplifying the integral.

In our exercise, integration by parts was necessary to integrate terms such as \(\int y^2 \ln(x) dx\) and \(\int \ln(x) dx\). In practice, choosing \(u = \ln(x)\) and \(dv = dx\) simplifies the product, allowing us to find the antiderivatives. Successfully integrating by parts can be crucial for finding the solution to certain differential equations and is a powerful tool when dealing with complicated integrals.

General Solution of Differential Equations

The general solution of a differential equation represents a family of solutions encompassing all possible particular solutions, corresponding to different initial conditions or constants of integration. When we integrate a differential equation, the constant of integration, \( C \), encapsulates this infinite set of potential solutions.

In the exercise, after manipulating and integrating, we obtain a non-explicit general solution involving the constant \( C \). It can be quite challenging to isolate \( y \) explicitly in some cases, as evident from the resulting quartic equation. The general solution provides valuable information about the behavior of the entire family of solutions. When sketching these curves, we can represent different members by varying the constant \( C \) within the solution expression, thereby visualizing an infinite number of possible graphs that satisfy the original differential equation.