Use Frobenious method to find the series solution about  of the equation

To solve the given differential equation using the Frobenius method, we first assume a solution of the form \( y = \sum_{n=0}^{\infty} a_nx^{n+r} \), where \( a_n \) are constants to be determined and \( r \) is a constant to be determined as well. Then, we substitute this expression into the differential equation and solve for \( r \) and the coefficients \( a_n \).

Given the differential equation:

\[ x(1-x)\frac{d^2y}{dx^2}-(1+3x)\frac{dy}{dx}-y = 0 \]

Let's substitute \( y = \sum_{n=0}^{\infty} a_nx^{n+r} \) into the differential equation:

\[ x(1-x)\frac{d^2}{dx^2} \left( \sum_{n=0}^{\infty} a_nx^{n+r} \right) - (1+3x)\frac{d}{dx}\left( \sum_{n=0}^{\infty} a_nx^{n+r} \right) - \sum_{n=0}^{\infty} a_nx^{n+r} = 0 \]

First, let's find the first and second derivatives:

\[ \frac{d}{dx} x^{n+r} = (n+r)x^{n+r-1} \]

\[ \frac{d^2}{dx^2} x^{n+r} = (n+r)(n+r-1)x^{n+r-2} \]

Now, we substitute these into the differential equation:

\[ x(1-x) \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r-2} - (1+3x) \sum_{n=0}^{\infty} (n+r)a_nx^{n+r-1} - \sum_{n=0}^{\infty} a_nx^{n+r} = 0 \]

\[ \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r} - \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r+1} - \sum_{n=0}^{\infty} (n+r)a_nx^{n+r} - 3 \sum_{n=0}^{\infty} (n+r)a_nx^{n+r} - \sum_{n=0}^{\infty} a_nx^{n+r} = 0 \]

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