Question

For the differential equation whose solution is $(x-h{)}^2+(y-k{)}^2=a^2$ (a is a constant), its

• A. order is 2
• B. order is 3
• C. degree is 2
• D. degree is 3

Answer 1

Answer: (A), (C)

We have $(x-h{)}^2+(y-k{)}^2=a^2.....$(1)
Differentiating w.r.t. $x$, we get
$2(x-h)+2(y-k)\frac{dy}{dx}=0$
$(x-h)+(y-k)\frac{dy}{dx}=\mathrm{0.....}$(2)
Differentiating w.r.t. $x$, we get
$1+{\left(\frac{dy}{dx}\right)}^2+(y-k)\frac{d^{2}y}{d{x}^2}=\mathrm{0.....}$(3)
From equation (3),
$y-k=-\left(\frac{1+p^2}{q}\right)$, where $p=\frac{dy}{dx},q=\frac{d^{2}y}{d{x}^2}$
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 ${\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^3=a^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^2$
which is the required differential equation