Q107 / 180 · KCET 2017 · Mathematics
General solution of differential equation \( \frac{d y}{d x}+y=1(y \neq 1) \) is
- Correct answerA\( \log \left|\frac{1}{1-y}\right|=x+C \)
- B\( \log |1-y|=x+C \)
- C\( \log |1+y|=x+C \)
- D\( \log \left|\frac{1}{1-y}\right|=-x+C \)
Solution
Given differential equation,
\[ \begin{array}{l} \frac{d y}{d x}+y=1(y \neq 1) \\ \Rightarrow \frac{d y}{d x}=(1-y) \\ \Rightarrow \frac{d y}{(1-y)}=d x \end{array} \]
Integrating both the sides, we get
\[ \begin{array}{l} \int \frac{d y}{(1-y)}=\int d x \\ \Rightarrow-\log |1-y|=x+C \\ \Rightarrow \log \left|\frac{1}{1-y}\right|=x+C \end{array} \]
\[ \begin{array}{l} \frac{d y}{d x}+y=1(y \neq 1) \\ \Rightarrow \frac{d y}{d x}=(1-y) \\ \Rightarrow \frac{d y}{(1-y)}=d x \end{array} \]
Integrating both the sides, we get
\[ \begin{array}{l} \int \frac{d y}{(1-y)}=\int d x \\ \Rightarrow-\log |1-y|=x+C \\ \Rightarrow \log \left|\frac{1}{1-y}\right|=x+C \end{array} \]
Hence, the correct answer is option: 'A'.