Problem 35 - Differential Equations and Linear Algebra

Question

Determine which of the five types of differential equations we have studied the given equation falls into (see Table \(1.12 .1),\) and use an appropriate technique to find the general solution. $$\left(x^{2}-1\right)\left(y^{\prime}-1\right)+2 y=0$$

Step-by-Step Solution

Answer

The given differential equation is \(\left(x^{2}-1\right)\left(y^{\prime}-1\right)+2 y=0\), which doesn't fall into any of the five types of differential equations we studied. However, by integrating both sides with respect to x, we find the general solution to be \(y = -\frac{1}{2}\left(\frac{x^2-1}{2} + \frac{1}{2}\ln{|x^2-1|}\right) + x + C\).

Step 1: Identify type of the differential equation

Given differential equation is: \(\left(x^{2}-1\right)\left(y^{\prime}-1\right)+2 y=0\) It's not easy to see directly which type of differential equation it is. We can rewrite the equation to get better insight: \(y^{\prime}-1 = -\frac{2y}{x^2 - 1}\) Now we can add 1 to both sides: \(y^{\prime} = -\frac{2y}{x^2 - 1} + 1\) Let's have a better look at this equation. Unfortunately, it's not one of the five types we learned so far. However, we can still try to solve it by another method. Let's inspect the problem further.

Step 2: Integrate with respect to x

We can try integrating the equation with respect to x to find the solution. Rearrange the equation to isolate \(y'\): \(y^{\prime}=-\frac{2y}{x^2-1}+1\) Now, let's integrate both sides with respect to x: \(\int 1 dy = -2 \int \frac{y}{x^2-1}dx + \int 1dx\)

Step 3: Solve the integrals

Now we need to solve the following integrals: - \(\int 1 dy\) - \(-2 \int \frac{y}{x^2-1}dx\) - \(\int 1dx\) The first and the third integrals are straightforward. We can use substitution to solve the second integral. Let \(u = x^2 - 1\) and \(du = 2xdx\). We can also solve for \(y\) in terms of \(u\): \(y = \frac{u+1}{2}\). The second integral now becomes: \(-2 \int \frac{y}{x^2-1}dx = -\int \frac{(\frac{u+1}{2})}{u} \frac{du}{2x}\) Solve the integrals to find: \(y = -\frac{1}{2}(\frac{u}{2}+\frac{1}{2}\ln{|u|}) + x + C\)

Step 4: Substitute back to get y in terms of x

Now we need to substitute back \(u = x^2 - 1\) to get \(y\) in terms of \(x\): \(y = -\frac{1}{2}\left(\frac{x^2-1}{2} + \frac{1}{2}\ln{|x^2-1|}\right) + x + C\) So, we have found the general solution for the given equation: \(y = -\frac{1}{2}\left(\frac{x^2-1}{2} + \frac{1}{2}\ln{|x^2-1|}\right) + x + C\)

Key Concepts

General Solution

When dealing with differential equations, finding the general solution is key. This solution encompasses all possible particular solutions of the differential equation. It contains arbitrary constants, which are adjusted based on initial or boundary conditions to find specific solutions. Here’s why understanding this concept is crucial:

• **Insight into Dynamics:** It provides a broad look at the behavior of the system described by the equation.
• **Basis for Specific Solutions:** By applying specific conditions, you can derive solutions that model real-world phenomena.
• **Comprehensive Understanding:** It allows mathematicians and scientists to predict numerous outcomes for varying conditions.

Clearly defining the general solution is a fundamental step in handling differential equations effectively. The arbitrary constant(s) present allow for flexibility, accommodating a wide range of situations.

Integration

Integration is a powerful tool often used to solve differential equations. By integrating a differential equation, we can move from the derivative back to the original function. This process involves finding antiderivatives, which are functions that produce the original integrand when differentiated.In our exercise, integration helps in solving the equation:

• **Direct Integration:** Simple integrals like \( \int 1 \, dx \) and \( \int 1 \, dy \) illustrate straightforward antiderivatives.
• **Complex Integrals:** For terms like \( -2 \int \frac{y}{x^2-1} \, dx \), integration can be trickier and requires techniques like substitution.
• **Solution Construction:** Using integration, we construct the solution back from its derivative form.

Integration not only aids in finding solutions but also enriches understanding by connecting different mathematical concepts.

Substitution Method

The substitution method is a strategic way to simplify integration, especially when dealing with more complex integrals. Substitution involves replacing a complicated expression with a new variable to make the integration process easier.Here’s a breakdown of how substitution simplifies our problem:

• **Variable Change:** Set \( u = x^2 - 1 \) to make the integrals more manageable.
• **New Expression:** Transform the original integrand into a simpler form, potentially leading to easier integration.
• **Reverse Substitution:** Once integrated, replace the substituted variable back into the original context.

Substitution is a powerful technique. It can turn difficult integrals into ones that are more approachable, making it a valuable tool in calculus and differential equations.