1.

Solution of the differential equation `(1+x^(2)) dy + 2xy dx = cot x dx ` isA. ` y = log |sin x | (1+x^(2))+C(1+x^(2))^(-1)`B. ` y = log |sin x|(1+x^(2))^(-1)+C(1+x^(2))`C. ` y = log |sin x | (1+x^(2))+C(1+x)`D. ` y = log |sin x| (1+x^(2))^(-1)`

Answer» Correct Answer - c
Given , `(1+x^(2))dy +2xy dx = cot x dx `
` rArr (1+x^(2)) dy = dx (cot x - 2xy ) `
` rArr (dy)/(dx) = (cot x - 2xy)/(1+x^(2))`
` rArr (dy)/(dx) +(2xy)/(1+x^(2))= (cot x)/(1+x^(2))`
Or comparing with the form `(dy)/(dx) +Py = Q` we get
` :. P = (2x)/(1+x^(2))and Q = (cot x)/(1+x^(2))`
` :. IF = e^(int Pdx) = e^(int (2x)/(1+x^(2) dx)) = e^(log(1+ x^(2))= 1+x^(2)`
The General solution of the given differential equation given by
`y * IF int Q xx IF dx +C`
` rArr (1+x)^(2) y = int( (1+x^(2))(cotx)/((1+x^(2)))) dx +C`
` rArr (1+x^(2)) y = log |sin x| +C`
` rArr y = log | sin x | (1+x^(2))^(-1)+C (1+x^(2))^(-1)`


Discussion

No Comment Found

Related InterviewSolutions