xy''-(x+1)y'+y=0

asked by guest
on Jan 15, 2025 at 9:40 am



You asked:

Investigate the equation: \(x \cdot y^{\prime \prime} - \left(x + 1\right) \cdot y^{\prime} + y = 0\).

MathBot Answer:

\[y{\left(x \right)} = \left(1 + x\right) C_{2} + x^{2} \left(1 + \frac{x}{3} + \frac{x^{2}}{12} + \frac{x^{3}}{60}\right) C_{1} + O\left(x^{6}\right)\]

\(C_{1}\), \(C_{2}\) are arbitrary constants.


\(O\) is the error term of the series expansion in big O notation. The error term describes the maximum size of the error near \(x = 0\) between the exact solution \(y{\left(x \right)}\), and the approximate solution \(x^{2} C_{1} \left(\frac{x^{3}}{60} + \frac{x^{2}}{12} + \frac{x}{3} + 1\right) + C_{2} \left(x + 1\right)\). \(y{\left(x \right)} = C_{2} \left(x + 1\right) + x^{2} C_{1} \left(\frac{x^{3}}{60} + \frac{x^{2}}{12} + \frac{x}{3} + 1\right) + O\left(x^{6}\right)\) means that there exists a positive constant \(M\) such that for \(x\) sufficiently close to zero \(|y{\left(x \right)} - \left(\left(1 + x\right) C_{2} + x^{2} \left(1 + \frac{x}{3} + \frac{x^{2}}{12} + \frac{x^{3}}{60}\right) C_{1}\right)| \le M |x^{6}|\).