<transcript><audio_8>get a 1 over minus I Omega mines s it's an additional man sign because you're on lower boundaries over our as plus I Omega you can make those two fractions</transcript><transcript><audio_9>one fraction by multiplying with s plus I got the first one and the other one is minus I Omega they get in the denominator and s square is omega squared and in the numerator now you gettin s plus I Omega minus s minus I</transcript><transcript><audio_10>Omega so - I Omega divided by 2 I it was an Omega so there you have your Laplace transform of the sine of Omega T and this works if the upper bounds of your integral vanish and for that you need to</transcript><transcript><audio_11>to solve the fridge question now using Laplace transforms so we take you double plus Omega square times u equals zero user of 0 cos 0 and nu prime sir it was Omega second order linear differential</transcript> get a 1 over minus I Omega mines s it's an additional man sign because you're on lower boundaries over our as plus I Omega you can make those two fractions one fraction by multiplying with s plus I got the first one and the other one is minus I Omega they get in the denominator and s square is omega squared and in the numerator now you gettin s plus I Omega minus s minus I Omega so - I Omega divided by 2 I it was an Omega so there you have your Laplace transform of the sine of Omega T and this works if the upper bounds of your integral vanish and for that you need to to solve the fridge question now using Laplace transforms so we take you double plus Omega square times u equals zero user of 0 cos 0 and nu prime sir it was Omega second order linear differential
{"source": "grass-yt-cc-by.-nmlvHEDweM|shard_10027_34|create_multimodal_data.extract_yt_and_generate_transcripts.22|shard_10027|create_multimodal_data.generate_captions.38|shard_10027|create_multimodal_data.generate_captions.38", "params": {"id": "-nmlvHEDweM", "channel_id": "UCEjpRpZSjy6mkwKqzeTeVrQ", "duration": 564, "view_count": 393, "age_limit": 0, "like_count": 1, "channel_follower_count": 4880, "license": "Creative Commons Attribution license (reuse allowed)", "uploader_id": "@MatheMagician", "yt_tar_file": "/leonardo_work/EUHPC_E03_068/audio/fixed_download/grass-yt-cc-by//shard_10027/-nmlvHEDweM.tar.gz"}}