Question

Solution of the differential equation (x-y)dy=(x+y+1)dx is _____

(a) \(  \sqrt{x^2+y^2} = e^{c tan^{-1}⁡\left(\frac{y+0.5}{x+0.5}\right)}\)

(b) \((x^2+y^2)^1.5 = c  cot^{-1} ⁡\left(\frac{y+0.5}{x+0.5}\right)\)

(c) \((x^2+y^2)^2 = e^{c cot⁡⁡\left(\frac{y+0.5}{x+0.5}\right)}\)

(d) \((x^2+y^2)^1 = c  tan⁡\left(\frac{y+0.5}{x+0.5}\right)\)

This question was addressed to me in an interview for job.

The question is from Reducible to Homogenous Form in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right choice is (a) \(  \sqrt{x^2+y^2} = e^{c tan^{-1}⁡\left(\frac{y+0.5}{x+0.5}\right)}\)

Easiest explanation: \(\frac{dy}{dx} = \frac{x+y+1}{x-y}\), by substituting x=X+h, y=Y+k

w.k.t (h,k) satisfies the equations as h+k+1=0 & h-k=0 –> h=k=-0.5

and hence the differential equation changes to the form \(\frac{dy}{dx}=\frac{X+Y}{X-Y} \) is a  homogenous equation thus put \(v = \frac{Y}{X} \rightarrow \,v + X \frac{dv}{dX} = \frac{1+v}{1-v}\)

\(X \frac{dv}{dX} = \frac{1+v^2}{1-v} \rightarrow \int \frac{1-v}{1+v^2} \,dv = ∫\frac{1}{X} \,dx = \int(\frac{1}{1+v^2} – \frac{v}{1+v^2}) \,dv = log X + c\)

tan^-1⁡v-0.5log (1+v^2) = log X+c –> \( tan^{-1} \frac{⁡Y}{X} – log⁡(\sqrt{X^2+Y^2}) = c\)

\(\sqrt{X^2+Y^2} = e^{c tan^{-1}⁡\left(\frac{y+0.5}{x+0.5}\right)}\)……from the equation X=x+0.5 & Y=y+0.5.