Solve : x log x (dy/dx) + y = (2/y) log x

1.	Solve : x log x (dy/dx) + y = (2/y) log x
Answer» Given differential equation may be written as (dy/dx) + (1/x log x)y = 2/x²(which is linear diff. equation) Here, P = 1/x log x and Q = 2/x² Now, ∫Pdx = ∫(1/x log x) dx [put log x = z, (1/x) dx = dz] = ∫(1/z) dz = log z = log(log x) I..F = e^∫Pdx= e^{log(log x)} = log x Hence, solution is given by y x log x = ∫(2/x²).log x dx = 2∫x^-2.log x dx = 2[(log x).(x^{-2 + 1})/(-2 + 1) - ∫(1/x)(-1/x)] dx = 2[(-1/x)log x + ∫(1/x²) dx] = 2[(-1/x)log x - (1/x)] + k = (-2/x)(log x) + k

Answer»

Given differential equation may be written as

(dy/dx) + (1/x log x)y = 2/x²(which is linear diff. equation)

Here, P = 1/x log x and Q = 2/x²

Now, ∫Pdx = ∫(1/x log x) dx [put log x = z, (1/x) dx = dz]

= ∫(1/z) dz = log z = log(log x)

I..F = e^∫Pdx= e^{log(log x)} = log x

Hence, solution is given by

y x log x = ∫(2/x²).log x dx = 2∫x^-2.log x dx

= 2[(log x).(x^{-2 + 1})/(-2 + 1) - ∫(1/x)(-1/x)] dx

= 2[(-1/x)log x + ∫(1/x²) dx]

= 2[(-1/x)log x - (1/x)] + k = (-2/x)(log x) + k

Discussion