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
Numerical Value
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Answer:
Correct Answer: 14. Differential Equation: $\frac{d^{2} y}{d x^{2}}$ Curves : $x+y=2, x y=1$
Solution:
- Given, $y^{2}=2 c(x+\sqrt{c})$
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
$$ \begin{array}{rlrl} y^{2} & =2 y \frac{d y}{d x} \quad x+\sqrt{y \frac{d y}{d x}} \\ \Rightarrow \quad y & =2 \frac{d y}{d x} \cdot x+2 y^{1 / 2} \frac{d y}{d x}{ }^{3 / 2} \\ \Rightarrow \quad y-2 x \frac{d y}{d x} & =2 \sqrt{y} \frac{d y}{d x}^{3 / 2} \\ \Rightarrow \quad & & \\ \Rightarrow \quad & & =4 y \frac{d y}{d x}^{3} \end{array} $$
Therefore, order of this differential equation is 1 and degree is 3.
