Question

Find the differential equation of (x-h)²+ (y-k)² = a²​

Answer 1

We need to find the differential equation of,

$\longrightarrow(x-h)^2+(y-k)^2=a^2\quad\quad\dots(1)$

Differentiating the equation wrt $x$

$\longrightarrow2(x-h)+2(y-k)y'=0$

$\longrightarrow x-h=-(y-k)y'$

Putting this value of $x-h$ in (1),

$\longrightarrow(y-k)^2(y')^2+(y-k)^2=a^2$

$\longrightarrow(y-k)^2\left[1+(y')^2\right]=a^2\quad\quad\dots(2)$

Differentiating this equation,

$\longrightarrow2(y-k)\left[1+(y')^2\right]y'+2(y-k)^2y'y''=0$

$\longrightarrow y-k=-\dfrac{\left[1+(y')^2\right]}{y''}$

Putting this value of $y-k$ in (2),

$\longrightarrow\dfrac{\left[1+(y')^2\right]^3}{(y'')^2}=a^2$

Differentiating this equation,

$\longrightarrow\dfrac{6\left[1+(y')^2\right]^2y'(y'')^3-2\left[1+(y')^2\right]^3y''y'''}{(y'')^4}=0$

$\longrightarrow\underline{\underline{3y'(y'')^2-(y')^2y'''-y'''=0}}$

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: answer:-}}}}}$

$\sf{ {[1+ {(\frac{dy}{dX})}^{2} \brack}^{3} = {a}^{2} {( \frac{ {d}^{2}y }{d {x}^{2} } )}^{2} }$

◉GIVEN:-

$\sf{ {(x - h)}^{2} + {(y - k)}^{2} = {a}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:\:\:\:\:\: \: (1)$

$\longrightarrow \: \sf{2(x - h) + 2(y - k) \frac{dy}{dx} = 0}$

$\longrightarrow \sf{(x - h) + (y - k) \frac{dy}{dx} = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2)}$

Again Differentiating

$\sf{(y - k) = - \frac{1 + { (\frac{dy}{dx} )}^{2} }{ \frac{ {d}^{2}y }{d {x}^{2} } } }$

Putting in EQ(2) we get

$\sf{(x - h) = - (y - k) \frac{dy}{dx} }$

$\longrightarrow \sf{ \frac{1 + { [ (\frac{dy}{dx}) ] }^{2} \frac{dy}{dx} } { \frac{ {d}^{2}y }{d {x}^{2} } } }$

Putting in EQ(1)we get

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

$\longrightarrow \sf{ {[1 + { (\frac{dy}{dx}) }^{2}] }^{2} \: [ { (\frac{dy}{dx}) }^{2} + 1] = {a}^{2} {( \frac{ {d}^{2} y}{d {x}^{2} }) }^{2} }$

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