Mathematics - Differential Equation Question with Solution | TestHub

$y^2=a\left(x+\frac{\sqrt{a}}{2}\right)=ax+\frac{a^{3/2}}{2}$   $...\left(1\right)$

$\Rightarrow 2y{y}^{\text{'}}=a$

Put in equation $\left(1\right)$ we get, 

$y^2=\left(2y{y}^{\text{'}}\right)x+\frac{{\left(2y{y}^{\text{'}}\right)}^{3/2}}{2}$

$\left(y^2-2xy{y}^{\text{'}}\right)=\frac{{\left(2y{y}^{\text{'}}\right)}^{3/2}}{2}$

Squaring both the sides we get, 

${\left(y^2-2xy{y}^{\text{'}}\right)}^2=2y^3{\left(y^{\text{'}}\right)}^3$

$\therefore$ Order $=1$

Degree $=3$

Degree $-$ order $=3-1=2$