Question

Integrate
y/x dy/dx =
$\sqrt{1 + {x}^{2} + {y}^{2} + {x }^{2} \times {y}^{2} }$
​

Answer 1

Answer:

$\sf \sqrt{1+y^2} =\dfrac{(\sqrt{1+x^2}) ^3}{3} +C$

Step-by-step explanation:

Given:

$\sf \dfrac{y}{x} \: \dfrac{dy}{dx} =\sqrt{1+x^2+y^2+x^2\:y^2}$

To Find:

The solution of the given differential equation

Solution:

$\sf \dfrac{y}{x} \: \dfrac{dy}{dx} =\sqrt{1+x^2+y^2+x^2\:y^2}$

$\sf \dfrac{y}{x} \: \dfrac{dy}{dx} =\sqrt{1+x^2+y^2(1+x^2)}$

$\sf \dfrac{y}{x} \: \dfrac{dy}{dx} =\sqrt{(1+x^2)\times (1+\:y^2)}$

$\sf \dfrac{y}{x}\: \dfrac{dy}{dx} =\sqrt{1+x^2} \times \sqrt{1+y^2}$

Making it variable separable,

$\sf \dfrac{y}{\sqrt{1+y^2} } \:dy=x\times \sqrt{1+x^2} \:dx$

Integrating on both sides,

$\displaystyle \sf \int\limits {\dfrac{y}{\sqrt{1+y^2} } } \, dy=\int\limits x\:{\sqrt{1+x^2} } \, dx$

Now let us assume that 1 + y² = t²

Differentiating on both sides we get,

2y dy = 2t dt

y dy = t dt

Also assume that 1 + x² = p²

Differentiate on both sides,

2x dx = 2p dp

x dx = p dp

Substituting in the above integral we get,

$\displaystyle \sf\int\limits {\dfrac{t}{\sqrt{t} ^2} } \, dt =\int\limits {p\times \sqrt{p^2}} \, dp$

$\displaystyle \sf \int\limits {1} \, dt=\int\limits {p^2} \, dp$

$\sf t=\dfrac{p^3}{3} +C$

Substitute the value of t and p,

$\sf \sqrt{1+y^2} =\dfrac{(\sqrt{1+x^2}) ^3}{3} +C$

This is the solution of the given differential equation.