Need Help?

Subscribe to Differential Equation

Subscribe
  • Notes
  • Comments & Questions

Find the general solution:
$$x^{2} y^{\prime \prime}+x y^{\prime}+\left(X^{2}-\frac{1}{9}\right) y=0$$

$$v=\sqrt{1/9}=1 / 3$$

$$v \neq$$ integer

$$y=c_{1} J_{v} x+c_{2} J_{-v}(x)$$

$$y=c_{1} J_{1 / 3} x+c_{2} J_{-1 / 3}(x)$$

Where $$J_{v}=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n ! \Gamma(1-v+n)}\left(\frac{x}{2}\right)^{2 n-v} $$

Find the general solution:
$$x^{2} y^{\prime \prime}+x y^{\prime}+\left(X^{2}-1\right) y=0$$

$$y=c_{1} J_{1}(x)+c_{2} y_{1}(x)$$

Where $$J v=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n ! \Gamma(1-v+n)}\left(\frac{x}{2}\right)^{2 n-v}$$

$$y_{v}=\frac{\cos (v \pi) J_{v}(x)-J_{-v}(x)}{\sin (v \pi)} $$

Find the general solution:
$$4 x^{2} y^{\prime \prime}+4 x y^{\prime}+\left(4 X^{2}-25\right) y=0$$

$$\div 4$$

$$x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{25}{4}\right) y=0$$

$$V=\sqrt{\frac{25}{4}}=\frac{5}{2} $$

$$y=c_{1} J_{5 / 2}+c_{2} J_{-5/2} $$

No comments yet

Join the conversation

Join Notatee Today!