Question

The solution of the differential equation
` x^(2) (dy)/(dx)= x^(2) + xy + y^(2) `
A. ` tan^(-1)(y/x)=2 log x+C`
B. ` tan^(-1)(y/x)=3 log x+C`
C. ` tan^(-1)(y/x)= log x +C`
D. ` tan^(-1)(y/x)= 4 log x+C`

Answer 1

Correct Answer - b
Given , ` x^(2) (dy)/(dx) = x^(2) + xy + y^(2) `
` rArr (dy)/(dx) = (x^(2) + xy + y^(2))/(x^(2))" "`…(i)
This is a homogenous differential equation . To simplify it , put y = vx.
` rArr (dy)/(dx) = v+ x (dy)/(dx) `
On substituting the values of y and `(dy)/(dx) ` in Eq. (i) we get
` v + x (dv)/(dx) = (x^(2)+x^(2)v+x^(2)v^(2))/(x^(2)`
` rArr v + x (dv)/(dx) = (x^(2)[ 1+ v+v^(2)])/(x^(2))`
` rArr v + x (dv)/(dx) = 1+ v + v^(2) - v `
` rArr x (dv)/(dx) = 1+v^(2) rArr 1/(1+v^(2))dv = 1/x dx`
On integrating sides , we get ` int (dv)/(1+v^(2))= int 1/x dx `
`rArr tan^(-1) v = log x + C `