The differential equation whose solution is (x -h)^2 + (y -k)^2 =… - Vidyadip

MCQ

The differential equation whose solution is $(x -h)^2 + (y -k)^2 = a^2$ is ($a$ is a constant)

A
${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {a^2}\frac{{{d^2}y}}{{d{x^2}}}$
✓
${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {a^2}{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}$
C
${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}}} \right]^3} = {a^2}{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}$
D
None of these

✓

Answer

Correct option: B.

${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {a^2}{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}$

b
Given, $(x-h)^{2}+(y-k)^{2}=a^{2}$

$\Rightarrow 2(x-h)+2(y-k) \frac{d y}{d x}=0$

$\Rightarrow(x-h)+(y-k) \frac{d y}{d x}=0 \quad \ldots$ (ii)

Again differentiating $(y-k)=-\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}$

Putting in Eq. (ii), we get $x-h=-(y-k) \frac{d y}{d x}$

$=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right] \frac{d y}{d x}}{\frac{d^{2} y}{d x^{2}}}$

Putting in Eq. (i), we get $=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left(\frac{d y}{d x}\right)^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}+\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}=a^{2}$

$\Rightarrow\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left[\left(\frac{d y}{d x}\right)^{2}+1\right]=a^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$

$\Rightarrow\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=a^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$

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