# Find the differential equation of the family of curves color(white)(.)x^2+y^2+2ax+2by+c=0,where a,b,c are parameters?

Jul 2, 2018

$\left(y + b\right) \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} + 1 = 0$

#### Explanation:

Differentiating ${x}^{2} + {y}^{2} + 2 a x + 2 b y + c = 0$

$2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 a + 2 b \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

or $x + y \frac{\mathrm{dy}}{\mathrm{dx}} + a + b \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ or $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x + a}{y + b}$

or $1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} + y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + b \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

i.e. $1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} + \left(y + b\right) \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

or $\left(y + b\right) \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} + 1 = 0$

Jul 3, 2018

${y}_{x x x} + {y}_{x}^{2} {y}_{x x x} - 3 {y}_{x} {y}_{x x}^{2} = 0$

#### Explanation:

${x}^{2} + {y}^{2} + 2 a x + 2 b y + c = 0$

Taking 1st, 2nd, 3rd derivatives:

1st: $q \quad 2 x + 2 y {y}_{x} + 2 a + 2 b {y}_{x} = 0$

2nd: $q \quad 2 + 2 {y}_{x}^{2} + 2 y {y}_{x x} + 2 b {y}_{x x} = 0 q \quad \square$

3rd: $q \quad 6 {y}_{x} {y}_{x x} + 2 y {y}_{x x x} + 2 b {y}_{x x x} = 0 q \quad \triangle$

• $\square \cdot {y}_{x x x} - \triangle \cdot {y}_{x x}$ to eliminate remaining $b$ reference

$\implies 2 {y}_{x x x} + 2 {y}_{x}^{2} {y}_{x x x} + \cancel{2 y {y}_{x x} {y}_{x x x}} - 6 {y}_{x} {y}_{x x}^{2} - \cancel{2 y {y}_{x x} {y}_{x x x}} = 0$

$\therefore {y}_{x x x} + {y}_{x}^{2} {y}_{x x x} - 3 {y}_{x} {y}_{x x}^{2} = 0$