Differential Equations 1 Question 14
14.
The differential equation representing the family of curves $y^{2}=2 c(x+\sqrt{c})$, where $c$ is a positive parameter, is of
$(1999,3 M)$
(a) order 1
(b) order 2
(c) degree 3
(d) degree 4
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Answer:
Correct Answer: 14. (a,c)
Solution:
- Given, $y^{2}=2 c(x+\sqrt{c})$ $\quad$ …….(i)
On differentiating w.r.t. $x$, we get
$ 2 y \frac{d y}{d x}=2 c \Rightarrow c=y \frac{d y}{d x} $
On putting this value of $c$ in Eq. (i), we get
$y^{2} =2 y \frac{d y}{d x} \quad (x+\sqrt{y \frac{d y}{d x}} )$
$\Rightarrow \quad y =2 \frac{d y}{d x} \cdot x+2 y^{1 / 2} (\frac{d y}{d x}){ }^{3 / 2} $
$\Rightarrow \quad y-2 x \frac{d y}{d x} =2 \sqrt{y} (\frac{d y}{d x})^{3 / 2} $
$\Rightarrow \quad (y-2 x \frac{d y}{d x})^2 =4 y (\frac{d y}{d x})^{3}$
Therefore, order of this differential equation is $1$ and degree is $3$.
