The discussion revolves around solving the differential equation \( xy' = y \) using the Frobenius series method. Participants seek clarification on the method and its application, particularly in relation to initial conditions and the analytic nature of the solution.
Participants express varying levels of understanding regarding the Frobenius method and its application. There is no consensus on the clarity of the textbook explanation or the best approach to solving the differential equation.
Some participants reference the need for initial conditions and the analytic nature of the solution, but these concepts remain partially explored and not fully resolved within the discussion.
Individuals interested in differential equations, particularly those learning about series solutions and the Frobenius method, may find this discussion relevant.
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?