Question

2xydy - (x2 + y2 + 1)dx=0 is​

Answer 1

Answer:

$\\\tt y^2 = x^2 -1 + Cx$

Step-by-step explanation:

Rearrange:

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

Substitute:

Let $\\\tt y^2 = t$

Then differentiating both sides with respect to dx

$\\\tt 2y\dfrac{dy}{dx} = \dfrac{dt}{dx}\\\tt$

We have,

$\\\tt \dfrac{dt}{dx} - \dfrac{1}{x}(t) = x + \dfrac{1}{x}$

Integrate on form:

$\\\tt \dfrac{dy}{dx} +P(x)y = Q(x)$

Integrating factor (IF):

$\\\tt e^{\int - \dfrac{1}{x} } = e^{-log_ex} \\\tt e^{log_e(\frac{1}{x} )} = \dfrac{1}{x}$

Now, Using general form for integration:

$\\\tt I.F.y = \int I.F \ Q(x) dx$

$\\\tt \dfrac{1}{x} (t) = \int \dfrac{1}{x} (x + \dfrac{1}{x} ) dx \\\tt = \int (1+ \dfrac{1}{x^2} )dx\\ \\\tt \dfrac{1}{x} (t)= x -\dfrac{1}{x}+C \\\tt t = x^2 -1 + Cx$

Using value of t here:

$\\\tt y^2 = x^2 -1 + Cx$