For differential equation \( (e^{x}+e^{-x})dy=(e^{x}-e^{-x})dx \), then
Match List-I with List-II
| List-I | List-II | ||
| (Condition on Differential Equation) | (Solution) | ||
| (A) | \( x=0,\ y=\log_{e}2 \) | (I) | \( y=\log_{e}\left(\displaystyle\frac{e^{x}+e^{-x}}{2}\right)-e \) |
| (B) | \( x=0,\ y=e \) | (II) | \( y=\log_{e}\left(\displaystyle\frac{e^{x}+e^{-x}}{e+e^{-1}}\right) \) |
| (C) | \( x=1,\ y=0 \) | (III) | \( y=\log_{e}(e^{x}+e^{-x}) \) |
| (D) | \( x=0,\ y=-e \) | (IV) | \( y=\log_{e}\left(\displaystyle\frac{e^{x}+e^{-x}}{2}\right)+e \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
Separating the variables, the numerator is the derivative of the denominator:
\( \displaystyle dy=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx\ \)
\( \displaystyle \Rightarrow\ y=\log_{e}(e^{x}+e^{-x})+C \)
Each condition now fixes the constant \( C \).
(A) \( x=0,\ y=\log_{e}2 \)
This matches (III).
At \( x=0 \), \( \log_{e}2=\log_{e}2+C \) gives \( C=0 \), so \( y=\log_{e}(e^{x}+e^{-x}) \).
(B) \( x=0,\ y=e \)
This matches (IV).
\( e=\log_{e}2+C \), so \( C=e-\log_{e}2 \) and \( y=\log_{e}\frac{e^{x}+e^{-x}}{2}+e \).
(C) \( x=1,\ y=0 \)
This matches (II).
\( 0=\log_{e}(e+e^{-1})+C \), so \( y=\log_{e}\frac{e^{x}+e^{-x}}{e+e^{-1}} \).
(D) \( x=0,\ y=-e \)
This matches (I).
\( -e=\log_{e}2+C \), so \( y=\log_{e}\frac{e^{x}+e^{-x}}{2}-e \).
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).