KCET · Maths · Differentiation
General solution of differential equation \( \frac{d y}{d x}+y=1(y \neq 1) \) is
- A \( \log \left|\frac{1}{1-y}\right|=x+C \)
- B \( \log |1-y|=x+C \)
- C \( \log |1+y|=x+C \)
- D \( \log \left|\frac{1}{1-y}\right|=-x+C \)
Answer & Solution
Correct Answer
(A) \( \log \left|\frac{1}{1-y}\right|=x+C \)
Step-by-step Solution
Detailed explanation
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
\[
\begin{array}{l}
\int \frac{d y}{(1-y)}=\int d x \\
\Rightarrow-\log |1-y|=x+C \\
\Rightarrow \log \left|\frac{1}{1-y}\right|=x+C
\end{array}
\]
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