Question

Solve the following differential equation :

√(1+x²+y²+x²y²) +xydy/dx=0

Answer 1

Given,

\(\sqrt{1 + x^2 + y^2 + x^2y^2} + xy \frac{dy}{dx} = 0\)

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

Integrating both sides, we get

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

Substitute, 1 + y² = t²  ⇒  2ydy = 2tdt

⇒ \(\int \frac{\sqrt{1 + x^2}}x dx= - \int dt \)

\(= -t + c\)

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

Now substitute, 1 + x² = k²

⇒ 2xdx = 2kdk

⇒ \(\int \frac{k^2}{x^2}dk = - \sqrt{1 + y^2} +c\)

⇒ \(\int \frac{k^2}{k^2-1}dk = - \sqrt{1 + y^2} +c\)

⇒ \(\int\left[\frac{k^2 - 1}{k^2 - 1}+ \frac 1{k^2 - 1}\right]dk = - \sqrt{1 + y^2} +c\)

⇒ \(k + \frac 12 \log\frac{k - 1}{k + 1} = -\sqrt{1 + y^2}+ c\)

Since, \(\int \frac 1{x^2 - d^2}dx = \frac 1{2d}\log\frac{x - d}{x + d}\) + constant

\(\therefore \sqrt{1 + x^2}+ \frac 12 \log\frac{\sqrt{1 + x^2}-1}{\sqrt{1 + x^2}+ 1} + \sqrt{1+ y^2} = C\)

Answer 2

Given,√(1+x²+y²+x²y²) +xydy/dx=0

By simplifying the equation, we get

Now substituting these value of t and m, we get