Question 5 from your assignment asks to find the second solution of the differential equation:

$y'' + \frac{2x}{1-x^2}y' - \frac{2}{1-x^2}y = 0, \quad \text{for } -1 < x < 1$ Given that one solution is $y_1 = x$.

Step-by-Step Solution:

This is a second-order linear homogeneous differential equation. To find the second solution $y_2(x)$, we will use the reduction of order method since we already know one solution $y_1 = x$.

Step 1: Setup Reduction of Order

Let the second solution $y_2$ be of the form: $y_2 = v(x)y_1 = v(x) \cdot x$ where $v(x)$ is a function to be determined. Substituting this into the differential equation requires computing the derivatives of $y_2$.

First derivative: $y_2' = v'(x) \cdot x + v(x)$

Second derivative: $y_2'' = v''(x) \cdot x + 2v'(x)$

Step 2: Substitute into the Original Equation

Now substitute $y_2$, $y_2'$, and $y_2''$ into the given differential equation:

$\left(v''(x) \cdot x + 2v'(x)\right) + \frac{2x}{1-x^2}\left(v'(x) \cdot x + v(x)\right) - \frac{2}{1-x^2}\left(v(x) \cdot x\right) = 0$

Simplifying this step by step:

1. Expand the terms: $v''(x) \cdot x + 2v'(x) + \frac{2x}{1-x^2}v'(x) \cdot x + \frac{2x}{1-x^2}v(x) - \frac{2x}{1-x^2}v(x) = 0$

2. The terms $\frac{2x}{1-x^2}v(x) - \frac{2x}{1-x^2}v(x)$ cancel out, so we are left with: $v''(x) \cdot x + 2v'(x) + \frac{2x^2}{1-x^2}v'(x) = 0$

Step 3: Simplify the Equation

Factor out $v'(x)$: $v''(x) \cdot x + v'(x) \left(2 + \frac{2x^2}{1-x^2}\right) = 0$

Simplify the term inside the parentheses: $2 + \frac{2x^2}{1-x^2} = \frac{2(1-x^2 + x^2)}{1-x^2} = \frac{2}{1-x^2}$

So the equation becomes: $v''(x) \cdot x + v'(x) \cdot \frac{2}{1-x^2} = 0$

Step 4: Solve the Equation for $v(x)$

This is a first-order equation for $v'(x)$. To solve, we treat this as a separable equation. Divide through by $v'(x)$ to separate variables:

$\frac{v''(x)}{v'(x)} = -\frac{2}{x(1-x^2)}$

Integrating both sides with respect to $x$, we get:

$v'(x) = \frac{1}{(1-x^2)}$

Now integrate again to find $v(x)$: $v(x) = \frac{1}{2} \ln\left|\frac{1+x}{1-x}\right| + C$

Step 5: Find the Second Solution $y_2(x)$

Now that we have $v(x)$, the second solution is: $y_2(x) = v(x) \cdot y_1 = x \cdot \left( \frac{1}{2} \ln\left|\frac{1+x}{1-x}\right| + C \right)$

Thus, the second solution to the differential equation is:

$y_2(x) = \frac{x}{2} \ln\left|\frac{1+x}{1-x}\right| + Cx$

Final Answer:

The second solution is $y_2(x) = \frac{x}{2} \ln\left|\frac{1+x}{1-x}\right|$, for $-1 < x < 1$.

Would you like any additional clarification or further details?

Here are 5 follow-up questions related to this:

1. How does the reduction of order method work for second-order differential equations?
2. Why is it necessary to use the known solution $y_1 = x$ in this problem?
3. Can the constant $C$ in the second solution affect the general solution?
4. What are the conditions for using the logarithmic function in this context?
5. How can the solution change if the domain for $x$ is extended beyond $-1 < x < 1$?

Tip: Reduction of order is a powerful method to find a second linearly independent solution when one solution is already known.