Question

What is the order of the differential equation whose solution is $x^2+y^2+2gx+2fy+c=0$?

Answer 1

$x^2+y^2+2gx+2fy+c=0$

$2x+2y{y}^{\prime }+2g+2f{y}^{\prime }=0-\left(i\right)$

Differentiating again

$2+2y{y}^{\prime }+2(y^{\prime }{)}^2+2f{y}^{\prime \prime }=0-(ii)$

Differentiating again

$2[y^{\prime }{y}^{\prime \prime }+y{y}^{\prime \prime \prime }+2y^{\prime }\underset{\left(iii\right)}{y}"+t{y}^{\prime \prime \prime }=0]$

$yy{"}^{\prime }+3y^{\prime }y"+t{y}^{\prime \prime \prime }=0-\left(iv\right)$

From equation (ii) finding value of $"t"$

$-2fy"=2+2yy"+yy"+2(y^{\prime }{)}^2$

$t=\underset{y^{\prime \prime }}{-1-yy"-y(y^{\prime }{)}^2}$

$y{y}^{\prime \prime \prime }+3y^{\prime }y"+\left(\underset{y"}{-1-yy"-(y^{\prime }{)}^2}\right)y^{\prime \prime \prime }=0$

$y{y}^{\prime \prime \prime }y"+3y^{\prime }(y"{)}^2-y^{\prime \prime \prime }-y{y}^{\prime \prime }{y}^{\prime \prime \prime }-(y^{\prime }{)}^{2}y{y}^{\prime \prime \prime }=0$

$-y^{\prime \prime \prime }(1+(y^{\prime }{)}^2)+3y^{\prime }(y^{\prime \prime }{)}^2=0$

$y^{\prime \prime \prime }(1+(y^{\prime }{)}^2)-3y^{\prime }(y^{\prime \prime }{)}^2=0$

It is a differential equation of order $3$.