Question

Form the differential equation of the family of curves represented by, $c{(y+c)}^2=x^3$ , where c is any arbitrary constant.

• A. $12y{\left(y^{{}^{\prime }}\right)}^2=x[8{\left(y^{{}^{\prime }}\right)}^3-27]$
• B. $12y{\left(y^{{}^{\prime }}\right)}^2=x[{\left(2y^{{}^{\prime }}\right)}^3-27]$
• C. $12y{\left(y^{{}^{\prime \prime }}\right)}^2=x[{\left(2y^{{}^{\prime }}\right)}^3-27]$
• D. $12y{\left(y^{{}^{\prime }}\right)}^2=x[8{\left(y^{{}^{\prime }}\right)}^3+27]$

Answer 1

Answer: (A), (B)

The correct options are
A $12y{\left(y^{{}^{\prime }}\right)}^2=x[8{\left(y^{{}^{\prime }}\right)}^3-27]$
B $12y{\left(y^{{}^{\prime }}\right)}^2=x[{\left(2y^{{}^{\prime }}\right)}^3-27]$

$c{(y+c)}^2=x^3....\left(i\right)$
Differentiating both sides w.r.t $x$
$2c(y+c)y^{\prime }=3x^2......\left(ii\right)$
$\frac{\left(ii\right)}{\left(i\right)}\Rightarrow \frac{2y^{\prime }}{y+c}=\frac{3}{x}\Rightarrow c=\frac{2}{3}{y}^{\prime }x-y$
Now using $\left(ii\right)$
$\Rightarrow 2y^{\prime }.(\frac{2}{3}{y}^{\prime }x-y).\frac{2}{3}{y}^{\prime }x=3x^2$
$\Rightarrow 4\left(y^{\prime }{)}^2\right(2y^{\prime }x-3y)=27x$
$\Rightarrow 12y(y^{\prime }{)}^2=x\left[8\right(y^{\prime }{)}^3-27]$
Option A and B are same.