🔵22 - Method of Variation of Parameters 1 - Non-Homogeneous Differential Equations

[@ahmadsainable3611]

Wow! I never expected to learn this easily from an African Man. Thank you for making everything easy and straightforward. Your channel worth millions of subscribers.❤

[@SkanCityAcademy_SirJohn]

great feeling isn't it? thanks so much...

[@jonathanmikitch3902]

Thank you for this playlist clearing up the whole class for me.

[@SkanCityAcademy_SirJohn]

Aww thanks so much. Where do you watch me from?

[@I-am-king.]

Well explained

[@SkanCityAcademy_SirJohn]

Thanks so much

[@HenockNahango]

Super well explained.....

[@SkanCityAcademy_SirJohn]

Thanks so so much

[@GTOnizvka]

Thanks, you are helping a lot, Hi from Canada!

[@SkanCityAcademy_SirJohn]

Aww, thanks so much. Keep watching. Greeting heard.

[@selaimaoka1807]

Thank you ❤

[@SkanCityAcademy_SirJohn]

Most welcome

[@kofiboahen2156]

Hello sir, your videos are super helpful. Please my question is, if y prime prime has a coefficient that is greater than 1, should I divide through all terms by the coefficient of the second derivative before I start?

[@SkanCityAcademy_SirJohn]

Thanks so much. No please, just solve like you do for any quadratic equation, at the end you will get the two required values of m, that is m1 and m2. Thanks for watching. Which campus do you watch from?

[@carultch]

You can if you want to, but you don't necessarily have to do that. Given the general case of: a*y" + b*y' + c*y = g(t) The homogeneous solution (aka complimentary solution) is still a linear combination of e^(r1*t) and e^(r2*t), where r1 and r2 are the solutions to the quadratic equation a*r^2 + b*r + c = 0 You can solve this one, with or without first dividing through by a. The solution to a*yh" + b*yh' + c*yh = 0, will be the same as the solution to: yh" + b/a*yh' + c/a*yh = 0 The particular solution will be: yp = yh1*integral -g(t)/(W*a) dt + yh2* integral g(t)/(a*W) dt where: yh1 and yh2 are the two components of the homogeneous solution W is the Wronskian of the two components of the homogeneous solution g(t) is the original non-homogeneous part of the diffEQ a, b, and c are the original constant coefficients on each y-term

[@carultch]

As an example, consider: 4*y" + y = tan(t/2) The homogeneous solution is a sine and cosine linear combination with a frequency of 1/2. yh = A*cos(t/2) + B*sin(t/2) The Wronskian is W=1/2. To construct our particular solution: yp = yh1*integral -yh2*g(t)/(a*W) dt + yh2*integral yh1*g(t)/(a*W) dt yp = cos(t/2)*integral -sin(t/2)*tan(t/2)/(4*1/2) dt + sin(t/2)*integral cos(t/2)/(4*1/2) dt Which simplifies as: yp = 1/2*cos(t/2)*integral -sin(t/2)*tan(t/2) dt + 1/2*sin(t/2)*integral cos(t/2)*tan(t/2) dt After carrying out these yp = -cos(t/2)*arctanh(sin(t/2)) Which means the solution is: y = A*cos(t/2) + B*sin(t/2) - cos(t/2)*arctanh(sin(t/2))

[@kevn3850]

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[@muhammadmustapha5067]

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[@muhammadmustapha5067]

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[@kevcopo]

Wow perfectly explained

[@SkanCityAcademy_SirJohn]

Thanks so so much

[@engineeringsurgery]

Please may we have Laplace and Fourier Series Lectures...

[@SkanCityAcademy_SirJohn]

Those will be available very soon please

[@Jeffrey237]

Very explainatory video 😊

[@SkanCityAcademy_SirJohn]

Thanks so much

[@nhonnguyenthi6335]

Can y1 be a constant?

[@alexanderakaih7436]

Please can you make a video on variation for systems

[@SkanCityAcademy_SirJohn]

Yes

[@XerathMTA]

6:33 does it matter which you call y1 and y2? because i initially had c1e^3x + c2e^2x and chose y1= e^3x and y2 = e^2x

[@SkanCityAcademy_SirJohn]

Don't please it doesn't not really matter. Only that the expression for yc should be the same as what I have there.

[@XerathMTA]

@@SkanCityAcademy_SirJohn thanks

[@SkanCityAcademy_SirJohn]

you are most welcome. Where do you watch me from?@@XerathMTA

[@JohnDoeTVChannel]

Shout out to you

[@SkanCityAcademy_SirJohn]

Thanks so much. Where do you watch from?

[@tabbsonsimukonda5329]

too good

[@orph2443]

can I use it always?

[@SkanCityAcademy_SirJohn]

Please be more specific with your question.

[@carultch]

You can use it for any diffEQ of the form: a*y" + b*y' + c*y = g(t) And you can also use it for diffEQ's of the form: y" + p(t)*y' + q(t)*y = g(t) as long as you know in advance, the two homogeneous solutions to this equation if g(t) were replaced with zero. Not always a straight forward process to find, as it is when your coefficients are just constants. The big challenge is whether or not these integrals are practical to solve or not. It is very hard to find integrals that simplify to elementary functions for this method, and that are still complicated enough that you'd have to use variation of parameters. I've typically seen secants and tangents as the g(t) function. Usually, when your g(t) is of one of the function families that "plays nicely" with differential equations, you can use other methods like undetermined coefficients or Laplace transforms, to solve them, and don't even need variation of parameters. These function families are the following: Polynomials of t Exponentials of t Sine and cosine functions of t Linear and multiplicative combinations of any of the above What these function families have in common, is that they can be annihilated by a "polynomial" of the differential operator. This means they will either diminish to zero, or loop back to themselves, when differentiated repeatedly.

[@assmptioncorrect198]

y1=xe^2x (because in que on RHS it already e^2x)