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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...Solve xy'= y using frobenius method
The explanation given in the book is very confusing can somebody explain in simple method.
Thanks
can you explain to me the first line?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 $\displaystyle y = \frac{y_{0}}{x_{0}}\ x$ ...
$\chi$ $\sigma$
Of course the first order DE...can you explain to me the first line?