$({e^x} + {e^{ - x}})dy - ({e^x} - {e^{ - x}})dx = 0$
$({e^x} + {e^{ - x}})dy - ({e^x} - {e^{ - x}})dx = 0$
Official Solution
VVidaara Team
✓ Verified solution
NCERT & Exemplar
.: We have $({e^x} + {e^{ - x}})dy - ({e^x} - {e^{ - x}})dx = 0$
$\Rightarrow dy = \cfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx$
…(1)
Integrating (1) both sides,
we get
$\int d y = \int {\cfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx} \Rightarrow y = \log |{e^x} + {e^{ - x}}| + C$
which is the required solution.
Community Answers (0)
Log in to post your own answer or join the discussion.
No comments yet — start the discussion.