The discussion focuses on solving the first-order differential equation \( xy' = y \) using the Frobenius series method. The solution is derived by assuming that \( y(x) \) is analytic at a point \( x_0 \) and expressed as a power series. The derivatives \( y^{(n)}(x_0) \) are calculated, leading to the conclusion that the solution is a linear function given by \( y = \frac{y_0}{x_0} x \). The method requires understanding the behavior of the function around the point \( x_0 \).
PREREQUISITESMathematicians, engineering students, and anyone interested in advanced methods for solving differential equations, particularly those using series expansions.
ssh said:Solve xy'= y using frobenius method
The explanation given in the book is very confusing can somebody explain in simple method.
Thanks
chisigma said:Suppose that the initial condition $\displaystyle y(x_{0})= y_{0}$ is given and that $y(x)$ is analytic in $x_{0}$ so that is...
$\displaystyle y(x)= \sum_{ n=0}^{\infty} \frac{y^{(n)} (x_{0})}{n!}\ (x-x_{0})^{n}$ (1)
The $\displaystyle y^{(n)} (x_{0})$ are computed as follows...
$\displaystyle y^{(1)} = \frac {y}{x} \implies y^{(1)}(x_{0}) = \frac{y_{0}}{x_{0}}$
$\displaystyle y^{(2)} = \frac{y^{(1)}}{x}- \frac{y}{x^{2}} \implies y^{(2)}(x_{0}) = \frac{y_{0}}{x_{0}^{2}} - \frac{y_{0}}{x_{0}^{2}}=0$
$\displaystyle y^{(3)} = -2\ \frac{y^{(1)}}{x^{3}} + 2\ \frac{y}{x^{3}} \implies y^{(3)}(x_{0}) = -2\ \frac{y_{0}}{x_{0}^{3}} + 2\ \frac{y_{0}}{x_{0}^{3}}=0$
... and so one. The solution is the linear funcion $\displaystyle y = \frac{y_{0}}{x_{0}}\ x$ ...
$\chi$ $\sigma$
dansingh said:can you explain to me the first line?