SATHEE: Differential Equations 1 Question 14

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

(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

$\begin{aligned} y^{2} & = 2 y \frac{d y}{d x} x + \sqrt{y \frac{d y}{d x}} \\ \Rightarrow y & = 2 \frac{d y}{d x} \cdot x + 2 y^{1 / 2} \frac{d y}{d x}^{3 / 2} \\ \Rightarrow y - 2 x \frac{d y}{d x} & = 2 \sqrt{y} {\frac{d y}{d x}}^{3 / 2} \\ \Rightarrow \\ \Rightarrow & = 4 y {\frac{d y}{d x}}^{3} \end{aligned}$

Therefore, order of this differential equation is 1 and degree is 3.

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Differential Equations 1 Question 14

14. The differential equation representing the family of curves y2=2c(x+c), where c is a positive parameter, is of

Numerical Value

Answer:

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14. The differential equation representing the family of curves $y^{2} = 2 c (x + \sqrt{c})$ , where $c$ is a positive parameter, is of