Answer to Question #178672 in Differential Equations for Tausif

Question

The difference equation generated by y=a/x+b

Answer 1

given: y = $\frac{a}{x}$ + b

differentiating y with respect to x we get,

y' = -$\frac{a}{x^2}$

again differentiating y' with respect to x we get,

y'' = $\frac{2a}{x^3}$

$\implies$ a = $\frac{y'' x^3}{2}$

putting this value of a in y' we get,

y' = -$\frac{1}{x^2}$ $\cdot$ $\frac{y'' x^3}{2}$

$\implies$ y' = -$\frac{y'' x}{2}$

$\implies$ 2y' + y''$\cdot$ x = 0

thus the above equation is the differential equation of given y.