Question:
If \(x\,dy = y\,dx + y^2\,dy\), \(y > 0\) and \(y(1) = 1\), then what is \(y(-3)\) equal to?
यदि \(x\,dy = y\,dx + y^2\,dy\), \(y > 0\) और \(y(1) = 1\), तो \(y(-3)\) किसके बराबर है?
Options:
- 3 only
- -1 only
- Both -1 and 3
- Neither -1 nor 3
- केवल 3
- केवल -1
- दोनों -1 और 3
- न तो -1 और न ही 3
Correct Answer: a)
Explanation:
Rearranging the given differential equation:
\[x\,dy – y\,dx = y^2\,dy\]
Dividing both sides by \(y^2\):
\[\frac{x\,dy – y\,dx}{y^2} = dy \implies -\left(\frac{y\,dx – x\,dy}{y^2}\right) = dy\]
\[-d\left(\frac{x}{y}\right) = dy \implies -\frac{x}{y} = y + C \implies \frac{x}{y} = -y – C\]
Given \(y(1) = 1\):
\[\frac{1}{1} = -1 – C \implies C = -2\]
So the relation is:
\[\frac{x}{y} = 2 – y \implies x = 2y – y^2 \implies y^2 – 2y + x = 0\]
For \(x = -3\):
\[y^2 – 2y – 3 = 0 \implies (y – 3)(y + 1) = 0 \implies y = 3 \text{ or } y = -1\]
Since it is given that \(y > 0\), we must discard \(y = -1\), giving \(y = 3\) only.
Hence, the correct option is (a).
अवकल समीकरण को व्यवस्थित करने पर:
\[x\,dy – y\,dx = y^2\,dy\]
दोनों पक्षों को \(y^2\) से विभाजित करने पर:
\[-\frac{y\,dx – x\,dy}{y^2} = dy \implies -d\left(\frac{x}{y}\right) = dy\]
समाकलन करने पर:
\[-\frac{x}{y} = y + C\]
दिया है \(y(1) = 1\):
\[-1 = 1 + C \implies C = -2\]
अतः समीकरण \(\frac{x}{y} = 2 – y \implies y^2 – 2y + x = 0\) है।
\(x = -3\) रखने पर:
\[y^2 – 2y – 3 = 0 \implies (y – 3)(y + 1) = 0 \implies y = 3 \text{ या } y = -1\]
चूँकि \(y > 0\) दिया गया है, इसलिए केवल \(y = 3\) मान्य होगा।
अतः सही विकल्प (a) है।