First Order Linear Differential Equations
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- เผยแพร่เมื่อ 19 ก.ย. 2024
- This calculus video tutorial explains provides a basic introduction into how to solve first order linear differential equations. First, you need to write the equation in standard form [y' + P(x)y = Q(x)] and then identify the functions P(x) and Q(x). Next, you need to determine the integrating factor I(x) using the formula I(x) = e^(integral of P(x)dx). Finally, you can use another formula to find the general solution of the first order linear differential equation y = 1/I(x) [Integral(I(x)Q(x)dx + C]. This video contains plenty of examples and practice problems on solving first order linear differential equations.
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Bernoulli's Equation for Differential Equations:
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Pls can you tell me what textbook you used for these, I think my lecturer is using same thing because the questions and solutions are all the same
I've been struggling with differential equations and in just a few minutes you made everything click. I honestly don't understand why university lecturers don't include a simple "and here's the actual steps you need to take to solve an equation like this" after all of the heavy theory.
exactly, theory pisses me off
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00:00 -how to
01:13 -prob1
04:28 -prob2
10:24 -prob3
18:58 -prob4
thank you
Thanks
13:41 - prob 4
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In the last example, you can simply separate the equation (x² - y)/x into two terms: x - y/x. Goes the same answer, sir.
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Professor Organic Chemistry Tutor, thank you for another fantastic video/lecture on First Order Linear Differential Equations. Introductory Differential Equations books use a different method to solve First Order Linear Differential Equations. Many Introductory Differential Equations books after finding the Integrating Factor, use it to multiply the standard form equation and then solve for y. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
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for the last problem, you can simply spit apart the fraction and make it x^/x and -y/x. then I added y/x to both sides and got dy/dx +y/x = x^2/x. This leaves you with a P(X)= 1/x and a Q(x)= x as well.
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This may have been mentioned before but at 10:06 14/2 =7, which then is x^3=x^3. Fantastic video! A great help after being away from my last calculus class for 20 years this semester and now taking DEs. Thank you for all the great content.
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For the last example there's a much quicker way to get it into standard form. Re-write (x^2-Y)/X as X^2/X - Y/X and the just add the Y/X term to the left side.
Repeating differential equations in uni and i swear the professor explains this stuff in the most convoluted way possible just to fuck with us. I'm honestly better off daydreaming during class and learning all this from youtube later than actually listening to her. Thank you for the explanation i can actually follow and implement
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