Question

Solution of the differential equation \(xy \frac{dy}{dx} = 1+x+y+xy\) is ______

(a) (y-x)-log(x(1+y))=c

(b) log(x(1+y))=c

(c) (y+x)-log(x)=c

(d) (y-x)-log(y(1+x))=c

This question was addressed to me by my school teacher while I was bunking the class.

The origin of the question is Separable and Homogeneous Equations topic in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct option is (a) (y-x)-log(x(1+y))=c

The explanation is: \(xy \frac{dy}{dx} = 1+x+y+xy\)

\(xy \frac{dy}{dx} = (1+x)+y(1+x)=(1+x)(1+y)\)

separating the variables & hence integrating

\(\frac{y}{1+y} \,dy = \frac{1+x}{x} \,dx\)

\(\int\frac{y}{1+y} \,dy = \int \frac{1+x}{x} \,x\)

\(\int\frac{(1+y)-1}{1+y} \,dy = \int\frac{1}{x} \,dx + \int1 \,dx\)

\(\int1 \,dy – \int\frac{1}{1+y} \,dy – log \,x – x = c\)

y – log(1+y) – log x – x = c

(y-x) – log(x(1+y)) = c is the solution.