Consider the second-order differential equation: $Consider the second-order differential equation: . Why is C<sub>0</sub> = 0 a regular singular point? A) The functions x^{2} \cdot \frac{7 x(x+1) }{5 x^{2}} and x \cdot\left(-\frac{7}{5 x^{2}}\right) both have convergent Taylor series expansions about 0 . B) The functions x \cdot \frac{7 x(x+1) }{5 x^{2}} and x^{2} \cdot\left(-\frac{7}{5 x^{2}}\right) both have convergent Taylor series expansions about 0 . C) \lim _{x \rightarrow 0} x \cdot\left(-\frac{7}{5 x^{2}}\right) =\infty D) \lim _{x \rightarrow 0} x \cdot\left(-\frac{7}{5 x^{2}}\right) \neq 0$ .
Why is C₀ = 0 a regular singular point?

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The Answer of Consider the second-order differential equation: . Why is C₀ = 0 a regular singular point? A) The functions $ x^{2} \cdot \frac{7 x(x+1)}{5 x^{2}} $ and $ x \cdot\left(-\frac{7}{5 x^{2}}\right) $ both have convergent Taylor series expansions about 0 . B) The functions $ x \cdot \frac{7 x(x+1)}{5 x^{2}} $ and $ x^{2} \cdot\left(-\frac{7}{5 x^{2}}\right) $ both have convergent Taylor series expansions about 0 . C) $ \lim _{x \rightarrow 0} x \cdot\left(-\frac{7}{5 x^{2}}\right)=\infty $ D) $ \lim _{x \rightarrow 0} x \cdot\left(-\frac{7}{5 x^{2}}\right) \neq 0 $

Consider the Second-Order Differential Equation x2⋅7x(x+1)5x2 x^{2} \cdot \frac{7 x(x+1)}{5 x^{2}} x2⋅5x27x(x+1)​

Consider the Second-Order Differential Equation $x^{2} \cdot \frac{7 x(x+1)}{5 x^{2}}$