Question:
What is the solution of the differential equation \(\frac{dx}{dy} = \frac{x+y+1}{x+y-1}\)?
अवकल समीकरण \(\frac{dx}{dy} = \frac{x+y+1}{x+y-1}\) का हल क्या है?
Options:
- \(y – x + 4\ln(x + y) = c\)
- \(y + x + c + \ln(x + y) = c\)
- \(y – x + \ln(x + y) = c\)
- \(y + x + 2\ln(x + y) = c\)
- \(y – x + 4\ln(x + y) = c\)
- \(y + x + c + \ln(x + y) = c\)
- \(y – x + \ln(x + y) = c\)
- \(y + x + 2\ln(x + y) = c\)
Correct Answer: c)
Explanation:
Let \(v = x + y\). Differentiating with respect to \(y\):
\[\frac{dv}{dy} = \frac{dx}{dy} + 1 \implies \frac{dx}{dy} = \frac{dv}{dy} – 1\]
Substituting into the equation:
\[\frac{dv}{dy} – 1 = \frac{v+1}{v-1} \implies \frac{dv}{dy} = \frac{v+1}{v-1} + 1 = \frac{2v}{v-1}\]
Separating variables:
\[\frac{v-1}{2v}\,dv = dy \implies \left(\frac{1}{2} – \frac{1}{2v}\right)dv = dy\]
Integrating both sides:
\[\frac{v}{2} – \frac{1}{2}\ln|v| = y + C’ \implies v – \ln|v| = 2y + C\]
Substitute back \(v = x + y\):
\[x + y – \ln(x + y) = 2y + C \implies y – x + \ln(x + y) = c\]
Hence, the correct option is (c).
माना \(v = x + y\)। \(y\) के सापेक्ष अवकलन करने पर:
\[\frac{dv}{dy} – 1 = \frac{v+1}{v-1} \implies \frac{dv}{dy} = \frac{2v}{v-1}\]
चरों को पृथक करने पर:
\[\frac{v-1}{2v}\,dv = dy \implies \frac{1}{2}\left(1 – \frac{1}{v}\right)dv = dy\]
समाकलन करने पर:
\[v – \ln|v| = 2y + C\]
\(v = x + y\) रखने पर:
\[y – x + \ln(x + y) = c\]
अतः सही विकल्प (c) है।