Question

Solution of the differential equation \(\frac{dy}{dx} + \frac{y}{x} = y^2\) x is ________

(a) \(\frac{1}{y} = -x + c\)

(b) \(\frac{1}{xy} = -x + c\)

(c) \(\frac{1}{(xy)^2} = -y + c\)

(d) \(\frac{1}{xy} = -x^2 + c\)

I had been asked this question in class test.

My question is taken from Bernoulli Equations in section Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right answer is (b) \(\frac{1}{xy} = -x + c\)

For explanation: Given equation is of the form \(\frac{dy}{dx} + Py = Qy^n\) …divide by y^2

 where P and Q are functions of x  hence this is Bernoulli’s equation in \(y, \frac{1}{y^2} \frac{dy}{dx} + \frac{1}{yx} = x\)

put \(\frac{1}{y} = t \rightarrow \frac{-1}{y^2} \frac{dy}{dx} = \frac{dt}{dx}\) substituting we get – \(\frac{dt}{dx} + \frac{t}{x} = x \,or\,  \frac{dt}{dx} – \frac{t}{x} = -x\)  

this equation is linear in t i.e it is of the form

\(\frac{dt}{dx} + Pt = Q, I.F = e^{\int P \,dx} =e^{\int \frac{-1}{x} \,dx} = e^{-log⁡x} = \frac{1}{x}\)  

 its solution is

\(te^{\int P \,dx} = \int Q e^{\int P \,dx} \,dx + c \rightarrow t \frac{1}{x} = \int -x * \frac{1}{x} \,dx + c \rightarrow \frac{t}{x} = -x + c \,but\, t = \frac{1}{y}\)  

                        thus solution is given by \(\frac{1}{xy} = -x + c\).