The solution of the differential equation `(dy)/(dx)=(a x+g)/(b y+f)` represents a circle when `a=b` b. `a=-b` c. `a=-2b` d. `a=2b`

Options

A. ` e^(x) - e^(-siny) + (x^(3))/3 = C`
B. ` e^(-x) e^(-siny) + (x^(3))/3 = C`
C. ` e^(x) +e^(-siny) + (x^(3))/3 = C`
D. ` e^(x) - e^(siny) - (x^(3))/3 = C`

Answer

c

Explanation

Given , ` cos y (dy)/(dx) = e^(x+siny) +x^(2)e^(siny)`
` rArr cos y (dy)/(dx) = e^(siny) (e^(x)+x^(2))`
` rArr int (cos y)/(e^(siny)) dy = int (e^(x)+x^(2))dx`
On integrating both sides , we get
` int (dt)/(e^(t))dx = int (e^(x)+x^(2))dx`
` rArr -e^(-t) = e^(x) +(x^(3))/3 -C`
` rArr e^(x) +e^(-siny) + (x^(3))/3 =C`