How can I solve this differential equation? : $xy \ dx-(x^2+1) \ dy = 0$

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Answer 1 · 2018-02-14T16:26:18

$y = Asqrt(x^2+1)$

we have in differential form:

$xy \ dx-(x^2+1) \ dy = 0$

If we put in standard form, and collect terms:

$1/yy \ dy/dx = x/(x^2+1)$

Which is a First Order Separable Ordinary Differential Equation, so we can separate the variables to get:

$int \ 1/y \ dy = int \ x/(x^2+1) \ dx$

We can manipulate the RHS integral as follows:

$int \ 1/y \ dy = 1/2 \ int \ (2x)/(x^2+1) \ dx$

And now both integrals are standard results, so integrating give us:

$ln|y| = 1/2ln|x^2+1| + C$

Noting that we require areal solution, and writing $C=lnA$ , we get:

$ln y = ln Asqrt(x^2+1)$

Giving us the General Solution:

$y = Asqrt(x^2+1)$

How can I solve this differential equation? : xy \ dx-(x^2+1) \ dy = 0 xy \ dx-(x^2+1) \ dy = 0