Thank you for reporting, we will resolve it shortly
Q.
For the differential equation whose solution is $(x-h)^2+(y-k)^2=a^2$ (a is a constant), its
Differential Equations
Solution:
We have $(x-h)^2+(y-k)^2=a^2.....$(1)
Differentiating w.r.t. $x$, we get
$ 2(x-h)+2(y-k) \frac{d y}{d x}=0 $
$ (x-h)+(y-k) \frac{d y}{d x}=0.....$(2)
Differentiating w.r.t. $x$, we get
$1+\left(\frac{d y}{d x}\right)^2+(y-k) \frac{d^2 y}{d x^2}=0.....$(3)
From equation (3),
$y-k=-\left(\frac{1+p^2}{q}\right)$, where $p=\frac{d y}{d x}, q=\frac{d^2 y}{d x^2}$
Putting the value of $y-k$ in equation (2), we get
$x-h=\frac{\left(1+p^2\right) p}{q}$
Substituting the values of $x-h$ and $y-k$ in equation (1), we get
$\left(\frac{1+p^2}{q}\right)^2\left(1+p^2\right)=a^2 $
or $\left[1+\left(\frac{d y}{d x}\right)^2\right]^3=a^2\left(\frac{d^2 y}{d x^2}\right)^2$
which is the required differential equation