Use Laplace Transform to solve IVP dy/dt+3y=13sin2t, y(0)=6 | DE
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- čas přidán 10. 09. 2024
- FYI: Access the table of Laplace Transform of some basic functions
bit.ly/3oHrpXu
Going backwards to get Inverse Laplace transforms
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I'm from México, i don't know what happened , but i can understand better in this video
I'm glad that it helps :)
Hi whats the diference if in the transform i have dy/dx?
tnks a lot.
Good question. In reality, there's no different. It's just a different way of writing the derivative. dy/dx means you differentiate equation y with respect to x. dy/dt means you differentiate equation y with respect to t. The main thing is that you need to recognize what the function/equation is and what it respects to. Hope it makes sense.
can you explain how 6/s^2+4 is broken up to. I know you take out the 6 but where does the 1/2 come from ?
What is the laplace transform of sin 2t. It is 2/(s^2 + 4), right? After you factored out 6, you left with 1 on the numerator, but when you took the L of sin2t, the numerator became 2. So to compensate it, you need to multiply that by 1/2. It's a little confusing to explain it in word. I hope it makes sense to you :)
Video bna rhi thi tmez ki awaz bhi lga leti.