SATHEE: Differential Equations 3 Question 4

Differential Equations 3 Question 4

4. A curve passes through the point $1, \frac{π}{6}$ . Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x} + \sec \frac{y}{x}, x > 0$ .

Then, the equation of the curve is

(2013 Adv.)

(a) $\sin \frac{y}{x} = \log x + \frac{1}{2}$

(b) $cosec \frac{y}{x} = \log x + 2$

(d) $\cos \frac{2 y}{x} = \log x + \frac{1}{2}$

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Answer:

Correct Answer: 4. (a)

Solution:

PLAN To solve homogeneous differential equation, i.e. substitute $\frac{y}{x} = v$

$∴ y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

Here, slope of the curve at $(x, y)$ is

$\begin{aligned} \frac{d y}{d x} = \frac{y}{x} + \sec \frac{y}{x} \\ Put \frac{y}{x} = v \\ ∴ v + x \frac{d v}{d x} = v + \sec (v) \Rightarrow x \frac{d v}{d x} = \sec (v) \\ \Rightarrow \int \frac{d v}{\sec v} = \int \frac{d x}{x} \Rightarrow \int \cos v d v = \int \frac{d x}{x} \\ \Rightarrow \sin v = \log x + \log c \Rightarrow \sin \frac{y}{x} = \log (c x) \end{aligned}$

As it passes through $1, \frac{π}{6} \Rightarrow \sin \frac{π}{6} = \log c$

$\begin{array}{ll} \Rightarrow & \log c = \frac{1}{2} \\ ∴ & \sin \frac{y}{x} = \log x + \frac{1}{2} \end{array}$

Differential Equations 3 Question 4

4. A curve passes through the point $1, \frac{π}{6}$ . Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x} + \sec \frac{y}{x}, x > 0$ .

Answer:

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Differential Equations 3 Question 4

4. A curve passes through the point 1,π6. Let the slope of the curve at each point (x,y) be yx+sec⁡yx,x>0.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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4. A curve passes through the point $1, \frac{π}{6}$ . Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x} + \sec \frac{y}{x}, x > 0$ .