I'm trying to think of how many ways there are to solve this. It's separable, as you demonstrated. You can also create an integrating factor to render it exact. Variation of parameters would work too. Are there any other techniques?
There are several of them, yes. You can solve it by Laplace transformation for instance or by my personal favorite, the Ansatz technique. For Ansatz you make a guess. If your guess works, it is the solution to the differential equation. It's almost cheating. Solve f'(x) = - f(x). Guess: let f(x) = A·e^( τ x). Then τ A·e^( τ x) = - A·e^( τ x). τ A·e^(τ x) + A·e^(τ x) = 0. τ + 1 = 0. τ = - 1. Therefore, f(x) = A·e^(- x) ◼
![y'' = (y')^2 [Autonomous Differential Equation]](http://i.ytimg.com/vi/csNVGS5cRW8/mqdefault.jpg)
Thank you so much for explaining in detail😊😊
+C on both sides explanation is 100% correct this time! Amazing video ❤
Hey sir. Can you please do a video on asymptotes next? I have an exam due next week and this topic is bothering me too much.
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How to sketch rational functions
Well done. Thank you.
I'm trying to think of how many ways there are to solve this. It's separable, as you demonstrated. You can also create an integrating factor to render it exact. Variation of parameters would work too. Are there any other techniques?
There are several of them, yes. You can solve it by Laplace transformation for instance or by my personal favorite, the Ansatz technique.
For Ansatz you make a guess. If your guess works, it is the solution to the differential equation. It's almost cheating.
Solve f'(x) = - f(x).
Guess: let f(x) = A·e^( τ x).
Then τ A·e^( τ x) = - A·e^( τ x).
τ A·e^(τ x) + A·e^(τ x) = 0.
τ + 1 = 0.
τ = - 1.
Therefore, f(x) = A·e^(- x) ◼
Haha, I see why it's your favorite 😍
Series?
@@PrimeNewtons Oh my god NO.
... maybe? It sounds interesting.
Please integrate ln(sin^2x)sin2x from 0 to pi/2
Make u = sin^2x
Then use the double angle formula
Sin^2x=½(1-cos2x)
Then use u-sub
@@PrimeNewtons i got 2ulnu-u. Then what is the value of ln(sin^2(0))