Question

Particular solution of the differential equation \(\frac{dy}{dx} = \frac{x+y}{x}:y(1)=1\) is _____

(a) x  =y log|x| + y

(b) y = y log|x| + 2x

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

(d) y = x log|x| + x

This question was posed to me in an online quiz.

I need to ask this question from First Order Linear Differential Equations topic in section Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct answer is (d) y = x log|x| + x

Explanation: \(\frac{dy}{dx} – \frac{y}{x} = 1\) is of the form \(\frac{dy}{dx}\)  + Py = Q where P & Q is a function of x only

given DE is linear DE in y here \(P=\frac{-1}{x}, Q=1, e^{\int P \,dx} = e^{\int \frac{-1}{x} \,dx} = e^{logx^{-1}} = \frac{1}{x}\)

Linear DE solution is given by \(ye^{\int P \,dx} = \int Q \,e^{\int P \,dx} dx + c \rightarrow y \frac{1}{x} = \int 1 \frac{1}{x} \,dx + c\)

\(\frac{y}{x}\) = log⁡|x|+c, given y(1)=1–>y=1 when x=1 i.e c=1 therefore its particular solution is given by y = x log|x| + x.