Answer:
To find the power series solution for the differential equation \(y'' - (x + 1)y' - y = 0\), we'll assume that the solution can be expressed as a power series of the form:
\[y(x) = \sum_{n=0}^{\infty} a_n x^n.\]
We'll substitute this power series into the given differential equation and solve for the coefficients \(a_n\). Let's go through the steps:
Step 1: Calculate the first and second derivatives of \(y(x)\):
\[y'(x) = \sum_{n=0}^{\infty} a_n n x^{n-1} = \sum_{n=1}^{\infty} a_n n x^{n-1},\]
\[y''(x) = \sum_{n=0}^{\infty} a_n n(n-1) x^{n-2} = \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2}.\]
Step 2: Substitute the derivatives into the differential equation:
\[y'' - (x + 1)y' - y = \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2} - (x + 1) \sum_{n=1}^{\infty} a_n n x^{n-1} - \sum_{n=0}^{\infty} a_n x^n = 0.\]
Step 3: Combine the series terms and set the coefficients of like powers of \(x\) to zero:
\[\sum_{n=2}^{\infty} a_n n(n-1) x^{n-2} - \sum_{n=1}^{\infty} a_n n x^n - \sum_{n=0}^{\infty} a_n x^n - \sum_{n=1}^{\infty} a_n n x^n = 0.\]
Rearranging the terms and grouping by the power of \(x\):
\[\sum_{n=2}^{\infty} a_n n(n-1) x^{n-2} - \sum_{n=1}^{\infty} a_n n(n+1) x^n - a_0 - \sum_{n=1}^{\infty} a_n n x^n = 0.\]
Step 4: Equate coefficients to zero:
For the terms with \(x^{n-2}\): \(a_n n(n-1) = 0\) for all \(n \geq 2\), which implies \(a_n = 0\) for \(n \geq 2\).
For the terms with \(x^n\) (\(n \geq 1\)): \(a_n n(n+1) + a_n n = 0\), which simplifies to \(a_n n(n + 1 + 1) = 0\). This implies \(a_n = 0\) for all \(n \geq 1\).
Step 5: Solve for \(a_0\):
From the constant term (\(x^0\)): \(-a_0 = 0\) implies \(a_0 = 0\).
In conclusion, all coefficients \(a_n\) are zero, except \(a_0\) which is also zero. This means the solution to the differential equation is the trivial solution \(y(x) = 0\).
Explanation: