Question

Solution of the differential equation (1 + y²) (1 + logx) dx + x dy = 0 is:

(a) \(\tan^{-1}y + \log|x|+ \frac{(\log|x|)^2}{2} = C\)

(b) \(\tan^{-1}y - \log|x|+ \frac{(\log|x|)^2}{2} = C\)

(c) \(\tan^{-1}y - \log|x|- \frac{(\log|x|)^2}{2} = C\)

(d) \(\tan^{-1}y + \log|x|- \frac{(\log|x|)^2}{2} = C\)

Answer 1

Correct option is (a) \( \tan^{-1}(y) + \log x+ \frac{(\log x)^2}2 = C\)

\((1 + y^2) (1+ \log x)dx + xdy = 0\)

\(\therefore (1 + y^2) (1 + \log x)dx = - xdy\)

\(\frac{(1 + \log x)}{x} dx = -\frac 1{1 + y^2}dy\)

Integrating both sides,

\(\therefore \int \left(\frac{1 + \log x}x\right)dx = - \int \frac 1{1 + y^2} dy\)

\(\therefore \int \frac 1 x dx + \int \frac{\log x}x dx = - \int \frac 1{1 + y^2} dy\)

Put log x = t

Differentiating wrt 'x'

\(\frac 1 x dx = dt\)

\(\therefore \int \frac 1x x + \int tdt = - \int \frac 1{1 + y^2}dy\)

\(\log x+ \frac{t^2}2 = - \tan^{-1}(y) + C\)

\(\log x+ \frac{(\log x)^2}2 = - \tan^{-1}(y) + C\)

\( \tan^{-1}(y) + \log x+ \frac{(\log x)^2}2 = C\)