Differential Equations 3 Question 4
4. A curve passes through the point $1, \frac{\pi}{6}$. 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) $\operatorname{cosec} \frac{y}{x}=\log x+2$
(c) $\sec \frac{2 y}{x}=\log x+2$
(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 $\frac{y}{x}=v$
$$ \therefore \quad y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} $$
Here, slope of the curve at $(x, y)$ is
$$ \begin{aligned} & \frac{d y}{d x}=\frac{y}{x}+\sec \frac{y}{x} \\ & \text { Put } \quad \frac{y}{x}=v \\ & \therefore \quad v+x \frac{d v}{d x}=v+\sec (v) \Rightarrow x \frac{d v}{d x}=\sec (v) \\ & \Rightarrow \quad \int \frac{d v}{\sec v}=\int \frac{d x}{x} \quad \Rightarrow \quad \int \cos v d v=\int \frac{d x}{x} \\ & \Rightarrow \quad \sin v=\log x+\log c \Rightarrow \quad \sin \frac{y}{x}=\log (c x) \end{aligned} $$
As it passes through $1, \frac{\pi}{6} \Rightarrow \sin \frac{\pi}{6}=\log c$
$$ \begin{array}{ll} \Rightarrow & \log c=\frac{1}{2} \\ \therefore & \sin \frac{y}{x}=\log x+\frac{1}{2} \end{array} $$
