Question

x(1+y2)dx-y(1+x2)dy=0

Answer 1

I hope u can understand

Answer 2

$\dfrac{1+x^{2} }{1+y^{2} } =c^{2}$, this is therequired solution of differential equation.

Step-by-step explanation:

The given differentia equation is:

$x(1+y^{2} )dx-y(1+x^{2} )dy=0$

⇒ $x(1+y^{2} )dx=y(1+x^{2} )dy$

⇒$\dfrac{x}{1+x^{2} } dx=\dfrac{y}{1+x^{y} } dy$

Integrating both sides, we get

$\int {\dfrac{x}{1+x^{2} }} \, dx =\int {\dfrac{y}{1+y^{2} }} \, dy$    .....(1)

Let $1+x^{2} =t$

$2xdx=dt$

Also,

$1+y^{2} =v2ydy=dv$

Equation(1) can be written as, we get

$\int {\dfrac{1}{2t }} \, dt =\int {\dfrac{1}{2v }} \, dv$

[ ∵ $\int {\frac{1}{x} } \, dx =\log x$]

$\dfrac{1}{2} \log t-\dfrac{1}{2} \log v=\log c$

Where, $\log c$ is called integration constant.

$\dfrac{1}{2} (\log t- \log v)=\log c$

$\log \dfrac{t}{v} =\log c^{2}$

$\dfrac{t}{v} =c^{2}$    ... (2)

Putting the vaue of t and v in equation (2), we get

$\dfrac{1+x^{2} }{1+y^{2} } =c^{2}$, this is therequired solution of differential equation.