Runge kutta method second order differential equation simple example(PART-1)

[@umerkhitab1954]

The video was very informative. One thing, there is a little mistake, while deriving g(x,y,z), both the terms with 1 are positive.

[@abdoyasser2155]

its the first time that i give an indian a like ....thx alot for the clear steps

[@sreeram3893]

Bro i have a small dout, in this g(x) dz/dx=1+2xy+x^2z

[@brennankohen6366]

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[@hayesabdullah2018]

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[@AbishekRajkumar2k]

Yes that should be “+ve”..

[@ganeshkharvi303]

Yes

[@kartikmathur3519]

Ultimate video Thank you for explaination Only mistake is, there will be +ve sign in g(x,y,z)

[@mathswithfarhan]

Buht achy Bhai 🎉🎉🎉l🎉🎉 Thank You So Much 😊☺️❤️

[@Amitkumar-ur2ly]

Nicely explained thanks a lot. It just cleared this complex topic into easy one 🙂🙂

[@EASYMATHSEASYTRICKS]

Most welcome 😊

[@dosenterkuat]

there is amistake in geq. g but your explanation is perfect. Thanks a lot bro

[@mitchieroa2164]

thanks to you I get to answer our final project

[@himanshu2471]

Won't the 1 on the RHS of the given problem be considered in the function while computing for consecutive values?

[@WillyChandra13]

Sir, why did you add minus on the second equation (dz/dx=all positive)? And why you multiply h two times on the K equation?

[@amirkhatami5229]

Thanks, significantly helpful,great job

[@gamechanger-tt9qh]

thank u sir first subscriber

[@johnvincentjimar4157]

Thank you sir, you made this topic easy, subbed

[@sagarrawal8332]

Dear sir, Is it necessary to calculate l4? what it is use for?

[@bpeet315]

Hello friend, l4 is used in case we would like to calculate over a range of x not just 0.1, it is used to calculate the new "z", just like the k's are used to calculate the new "y"

[@ozonewagle]

Thank you so much. But i have one doubt. Why are we not calculating L it similar like k. If we dont need L then why are we doing till end

[@shivanichiplunkar]

It helped a lot 👍 Thanks

[@ShortSweet546]

Thankful sir ,amezing video

[@EASYMATHSEASYTRICKS]

Most welcome

[@Jenny-ie2vj]

Thanks sir, this helps me very much

[@BIJINICB]

Thankyou for this video. 👍🏻

[@asifimtiaz6353]

why we consider only k terms to the ultimate solution? why not both the k and l terms? please someone kindly clear this topic.

[@samanpaydar4415]

Awesome thank you very much 😍

[@vinaydesai8104]

Can you tell me the book which contains this type of problems? btw nice explanation

[@harshittyagi1805]

Now better understand thank you so much 👍👍

[@EASYMATHSEASYTRICKS]

You are welcome!

[@mudobirahmed80]

Thanks a lot....the video helped a lot.

[@seaksothearith]

Why do we need to calculate L4?

[@sagarrawal8332]

I was also thinking same.

[@ertugrul-bektik]

If you want to find y(0.2) you will need z(0.1). And the equation: z(0.1) = z0 + h/2 *[ L1 + 2*L2 + 2*L3 + L4 ]

[@felixruiztorrez5192]

L4 is the k4 equivalent for the second equation. He rewrites the second order differential equation into a system of two first order differential equations.

[@krishnakadiyam2864]

can you please solve an example of two 2nd order coupled differential equation?

[@devarajkt8670]

Helo brother after equating the terms each term will be positive there is no negative in the equation plz check it in text book..... page no.80 of author dr.KSC

[@ellery9528]

Very good thank you💫

[@tamermohamed8652]

good effort can i ask for your help??

[@vamsinaidu8690]

M3 total all modules upload sir with names and numbers of modules ❤️❤️

[@EASYMATHSEASYTRICKS]

All modules uploaded

[@vamsinaidu8690]

@@EASYMATHSEASYTRICKS in which play list sir only Laplace and 5th module is there sir

[@Joshtube309]

Sir how to do Runge Kutta for 2nd Order non-linear?

[@saniyanazdandin4725]

Thank you so much sir for this videos By watching ur videos only I cleared my m1 andm4 in 1st attempt only

[@EASYMATHSEASYTRICKS]

thank you

[@muhammadbasheer925]

Sir can u do video on taylors series second order

[@shahidhussain3021]

Bhai ...gazab...tq bro

[@mahiboobjilani7303]

Thank you very much sir

[@rakshitatembadmani1753]

Sir hw to solve 2a/r=1+cos thetha and 2a/r=1-cos thetha.....to show tht two lines intersect orthagonally....????

[@geniusmathematics9123]

Divide both.. And prove theta equal to 90°

[@suhani8948]

(i) Solve the differential equation (dy)/(dx) = 2xy, y(0) = 1 by Rungekutta second order method, where 0≤x≤1. Sir can you solve this sum

[@samitkhanal1737]

thank you very effective

[@md.asifiqbal3301]

thanks a lot brother

[@bubblelay6516]

Sir I have a question regarding with 2nd order differential equation using 4th order Runge kutta method is there any way I can contact u?

[@IshithaN]

Thank you sir🙏

[@Lakshmipathy_b]

Sir solve more problem on module 2 or only 4 problem is sufficient and what about module 5 sir please reply for my comment please I depend on your videos only sir thank you so much for for ur effort sir 😍😍

[@EASYMATHSEASYTRICKS]

ok

[@mdkhali2946]

Thanks so much

[@PardeepSingh-wj8lc]

Very nice

[@khagendraphuyal3763]

y'(0.1) pani sodya bela k garne??

[@feetify2009]

Since we have supposed y'= dy/dx=z Calc. Z(0.1)=z0+1÷6( l1+2*l2 + 2*l3+l4)

[@solution8103]

Thanks sir

[@sabbirahmedsakib7526]

Thank you

[@saniyanazdandin4725]

Sir plz do remaining prblms of module 2

[@EASYMATHSEASYTRICKS]

ok

[@uttkarshsachan4063]

What about the second iteration.

[@ertugrul-bektik]

If you want to find y(0.2) you will need z(0.1). And the equation: z(0.1) = z0 + h/2 *[ L1 + 2*L2 + 2*L3 + L4 ]

[@uttkarshsachan4063]

@@ertugrul-bektik Thanks for the explaination.

[@laxmikanthbpujari1749]

ossum sir,,,very helpfull

[@farhanaaz8964]

Please i request you to add Series of legendry polynomial derivation

[@gayitrigudi6855]

Thank u somuch sir

[@Akshay-vi6rw]

How to solve this with Second order equation....

[@aseer2718]

Thanks.

[@maleka0069]

Tq u sir

[@AbishekRajkumar2k]

Correction for dz/dx = 1+x^2z + 2xy…

[@teja3532]

Tq sir I'll get esy tips from you

[@abhishekagarwal3852]

What is the purpose of L4

[@ertugrul-bektik]

If you want to find y(0.2) you will need z(0.1). And the equation: z(0.1) = z0 + h/2 *[ L1 + 2*L2 + 2*L3 + L4 ]

[@cagataybaseymez6684]

i understood, thank you.

[@RakeshYadav-vc8hw]

❤😊🙏

[@DJCandy9068]

Thank you

[@EASYMATHSEASYTRICKS]

You're welcome

[@kumarswamy818]

Tq sir