Question

) For the following differential equation locate and classify its singular points on the x-axis

 

i) equation

 

ii) equation

09 Jan 2026
Answer :
Word Count : 315
i) The differential equation is [ x^3(x-1)y'' - 2(x-1)y' + 3xy = 0 ] We write it in standard form by dividing through by the coefficient of (y''), i.e., (x^3(x-1)): [ y'' - \frac{2(x-1)}{x^3(x-1)} y' + \frac{3x}{x^3(x-1)} y = 0 ] Simplify the coefficients: [ y'' - \frac{2}{x^3} y' + \frac{3}{x^2(x-1)} y = 0 ] Here, (P(x) = -\frac{2}{x^3}) and (Q(x) = \frac{3}{x^2(x-1)}). Singular points occur where the coefficients are not analytic, i.e., where (x^3(x-1) = 0), giving (x = 0) and (x = 1). * At (x = 0): [ (x-0)P(x) = x\left(-\frac{2}{x^3}\right) = -\frac{2}{x^2} \quad \text{not analytic}, ] [ (x-0)^2 Q(x) = x^2 \cdot \frac{3}{x^2(x-1)}
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