CUET · MATHS · PYQ PAPER 2023
The solution of the differential equation \(\log(\frac{dy}{dx}) = ax + by\) is:
- A \(\frac{1}{b}e^{by} = \frac{1}{a}e^{ax} + C\) where C is an arbitrary constant
- B \(\frac{1}{b}e^{-by} = \frac{1}{a}e^{ax} + C\) where C is an arbitrary constant
- C \(-\frac{1}{b} e^{-b y}=\frac{1}{a} e^{a x}+C\) where C is an arbitrary constant
- D \(-\frac{1}{b} e^{b y}=\frac{1}{a} e^{a x}+C\) where C is an arbitrary constant
Answer & Solution
Correct Answer
(C) \(-\frac{1}{b} e^{-b y}=\frac{1}{a} e^{a x}+C\) where C is an arbitrary constant
Step-by-step Solution
Detailed explanation
\(\log(\frac{dy}{dx}) = ax + by\) \(\frac{dy}{dx} = e^{ax+by} = e^{ax} e^{by}\) \(e^{-by} dy = e^{ax} dx\) \(\int e^{-by} dy = \int e^{ax} dx\) \(-\frac{1}{b} e^{-by} = \frac{1}{a} e^{ax} + C\)
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