Question

If $l$ and $m$ are the degree and the order respectively of the differential equation of the family of all circles in the $XY$ plane with radius 5 units, then $2l+3m=$

• A. 5
• B. 10
• C. 15
• D. 7

Answer 1

Answer: (B)

Family of all circles in the $xy$ -planewith radius $5$ units with center $\left(x_1,y_1\right)$ is
${\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2=5^2$
$\Rightarrow {\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2=25...(i)$
differential w.r.t ' $x^{{}^{\prime }}$
$\Rightarrow 2\left(x-x_1\right)+2\left(y-y_1\right)y^{{}^{\prime }}=0\cdots (ii)$
$\Rightarrow \left(x-x_1\right)=-\left(y-y_1\right)y^{{}^{\prime }}$
Now Eq. (ii) diff again w.r.t ' $x^{{}^{\prime }}$,
we get
$2+2\left(y^{{}^{\prime \prime }}\left(y-y_1\right)+{\left(y^{{}^{\prime }}\right)}^2\right]=0$
$\Rightarrow \left(y-y_1\right)=\frac{1-{\left(y^{{}^{\prime }}\right)}^2}{y^{{}^{\prime \prime }}}$
Sub. values in Eq. (i)
$={\left[\frac{1+{\left(y^{{}^{\prime }}\right)}^2}{y^{{}^{\prime \prime }}}\cdot y^{{}^{\prime }}\right]}^2+\left[\frac{1+{\left(y^{{}^{\prime }}\right)}^2}{y^{{}^{\prime \prime }}}\right]=25$
$\Rightarrow \frac{\left(1+y^{{}^{\prime }2}\right)}{{\left(y^{{}^{\prime \prime }}\right)}^2}\cdot {\left(y^{{}^{\prime }}\right)}^2+\frac{1+{\left(y^{{}^{\prime }}\right)}^2}{{\left(y^{{}^{\prime \prime }}\right)}^2}=25$
$\Rightarrow \frac{{\left(1+y^{{}^{\prime }}\right)}^3}{{\left(y^{{}^{\prime \prime }}\right)}^2}=25$
$\Rightarrow 25{\left(y^{{}^{\prime \prime }}\right)}^2={\left[1+{\left(y^{{}^{\prime }}\right)}^2\right]}^3$
So, order $=2$ and degree $=2$
$\therefore l=2$ and $m=2$
Now, $2l+3m=2\times 2+3\times 2$
$=4+6=10$