Solve the differential equations by the Frobenius method; observe that you get only one solution. (Note, also, that the two values of \(s\) are equal or differ by an integer, and in the latter case the larger \(s\) gives the one solution.) Show that the conditions of Fuchs's theorem are satisfied. Knowing that the second solution is \(\ln x\) times the solution you have, plus another Frobenius series, find the second solution. $$4 x^{2}(x+1) y^{\prime \prime}-4 x^{2} y^{\prime}+(3 x+1) y=0$$

Second-order linear differential equations are equations involving a function, its first derivative, and its second derivative. These equations are significant in various fields of science and engineering. The general form looks like this: \[ a(x) y'' + b(x) y' + c(x) y = d(x). \]\Here, the functions \(a(x)\), \(b(x)\), \(c(x)\), and \(d(x)\) can vary, but most commonly, \(d(x) = 0\), making the equation homogeneous. In our example equation is: \[ 4 x^{2}(x+1) y'' - 4 x^{2} y' + (3 x + 1) y = 0. \]To solve these, you typically bring the equation into a standard form by dividing all terms by \(a(x)\). In our case: \[ y'' - \frac{1}{x+1} y' + \frac{3x + 1}{4x^2(x+1)} y = 0. \]\br\ Understanding how to manipulate these coefficients is key to solving and understanding second-order linear differential equations.

The indicial equation arises when we apply the Frobenius method to solve differential equations. This method assumes a power series solution of the form \(y = x^s\).\br\ When we plug this assumed solution into the differential equation, we look at the lowest power of \(x\) to form the indicial equation.\br\ For our example: \[ 4 x^{2}(x+1) y'' - 4 x^{2} y' + (3 x + 1) y = 0, \]\assuming \(y = x^s\), we substitute into the equation and find the indicial equation: \[ s(s-1) = 0. \]\br\ This gives the roots \( s = 0 \) and \( s = 1 \). Both roots are real and differ by an integer, so only one solution initially emerges.

Fuchs's theorem is essential in understanding the solutions of differential equations around singular points. It states that if a point is a regular singular point of a second-order linear differential equation, then you can use techniques like the Frobenius method to find solutions.\br\ Conditions of Fuchs's theorem include: \[ a(x), \]\br\ is a standard form where \(a(x) e 0\) but it must be expandable into a Laurent series around the singular point \(x = 0\).\br\ In our example: \[ 4 x^{2}(x + 1) y'' - 4 x^{2} y' + (3 x + 1) y = 0, \] The singularity is at \( x = 0 \). Because the equation is in Frobenius standard form and the coefficients are expandable as a series, it confirms that conditions of Fuchs's theorem are satisfied. This assures us that a Frobenius series solution is applicable.

A series solution involves expressing the solution of a differential equation as an infinite sum of terms (usually power series). For the Frobenius method, we assume a solution in the form of: \[ y = x^s \sum_{n=0}^{\infty} a_n x^n, \] where \(a_0 eq 0\).\br\ Substituting this into the differential equation, we equate coefficients of equal powers of \(x\) to solve for \(a_n\) terms.\br\ For our example equation: \[ 4 x^{2}(x+1) y'' - 4 x^{2} y' + (3 x + 1) y = 0, \] we determined the roots \( s = 0 \) and \( s = 1 \).\br\ For \(s = 1\), we get the primary solution \( y_1 = x \).\br\ To find the second solution, knowing it involves \( \text{ln} x \), we get: \[ y_2 = y_1 \text{ln} x + A x = x \text{ln} x. \] Combining these solutions gives us a comprehensive understanding of the behavior around the singularity.