Question

Solution of the differential equation \(\frac{dy}{dx} = \frac{y-x+1}{y+x+5}\) is ______

(a) \(π-tan{-1}⁡ \frac{y+3}{x+2} – log\sqrt{(x+2)^2+(y+3)^2}=c\)

(b) \(-tan{-1}⁡⁡ \frac{y+3}{x+2} – log\sqrt{(x+2)^2+(y+3)^2}=c\)

(c) \(tan{-1}⁡⁡ \frac{y+3}{x+2} – log\sqrt{x+y+5}=c\)

(d) \(-cot{-1}⁡⁡ \frac{y+3}{x+2} – log\sqrt{x+y+5}=c\)

I got this question in an international level competition.

I need to ask this question from Reducible to Homogenous Form topic in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right option is (b) \(-tan{-1}⁡⁡ \frac{y+3}{x+2} – log\sqrt{(x+2)^2+(y+3)^2}=c\)

To explain: \(\frac{dy}{dx} = \frac{y-x+1}{y+x+5}\) –> x=X+h, y=Y+k

h+k+5=0 & k-h+1=0 solving we get h=-2, k=-3

\(\frac{dY}{dX} = \frac{Y-X}{Y+X}\), put Y=vX we get an new equation

\(V + X\frac{dV}{dX} = \frac{v-1}{v+1} \rightarrow X\frac{dV}{dX} = \frac{-(1+v^2)}{1+v}\)

 or

\(-\int \frac{1+v}{1+v^2} dv = \int \frac{1}{X} dX = \int \frac{1}{1+v^2} + \frac{v}{1+v^2} \,dv\)

-tan^-1⁡v-0.5 log(1+v^2)=log x +c

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

\(-tan{-1}⁡⁡ \frac{y+3}{x+2} – log\sqrt{(x+2)^2+(y+3)^2}=c\) is the solution.