Question

The general solution of \[ \log\left(\frac{dy}{dx}\right)=ax+by \] is

• A. \(e^{ax}+e^{by}=k\)
• B. \(be^{ax}+ae^{-by}=k\)
• C. \(be^{-ax}+ae^{-by}=k\)
• D. \(e^{-ax}-e^{-by}=k\)

Hint:

For equations of the form \[ \boxed{\log\left(\frac{dy}{dx}\right)=f(x)+g(y),} \] first write \[ \boxed{\frac{dy}{dx}=e^{f(x)}e^{g(y)}} \] and then separate the variables.

Answer 1

The Correct Option is B

Step 1: Rewrite the differential equation. Given, \[ \log\left(\frac{dy}{dx}\right)=ax+by. \] Hence, \[ \frac{dy}{dx}=e^{ax+by} =e^{ax}e^{by}. \] Therefore, \[ e^{-by}\frac{dy}{dx}=e^{ax}. \]

Step 2: Separate and integrate. Since \[ \frac{d}{dy}(e^{-by}) =-be^{-by}, \] we have \[ -\frac1b\,d(e^{-by}) = e^{ax}\,dx. \] Integrating, \[ -\frac1b\,e^{-by} = \frac1a\,e^{ax}+C. \] Multiplying by \(ab\), \[ be^{ax}+ae^{-by}=k, \] where \(k\) is an arbitrary constant. Hence, \[ \boxed{(B)} \] is the correct answer.