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- Feb 13, 2012

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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

$\displaystyle y(x)= \sum_{ n=0}^{\infty} \frac{y^{(n)} (x_{0})}{n!}\ (x-x_{0})^{n}$ (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 $\displaystyle y = \frac{y_{0}}{x_{0}}\ x$ ...

$\chi$ $\sigma$

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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 $\displaystyle y = \frac{y_{0}}{x_{0}}\ x$ ...

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

Of course the first order DE...can you explain to me the first line?

$\displaystyle y^{\ '} = \frac{y}{x},\ y(x_{0}) = y_{0}\ (1)$

... can be solved separain the variables and its soltion is $\displaystyle y=c\ x$...if Youy want to use the Frobenious method however, You must hypotize that $y(x)$ is analitic in $x_{0}$...

Kind regards

$\chi$ $\sigma$

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