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Math E Connect @UClf9th82zFuj8ZyXUkeYK8Q@youtube.com

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Asst. Prof. Shafiqahmed M Yellur, Jain College of Engineeri


01:05
National Science Day-2024
02:01
Solve πŸ‘πŸ 𝝏𝒖/𝝏𝒕=(𝝏^𝟐 𝒖)/〖𝝏𝒙〗^𝟐 (𝟎, 𝟏) given that h=0.25, upto t=5 𝒖=(𝟎,𝒕)=𝟎=𝒖(𝟏,𝒕), 𝒖=(𝒙,𝟎).
02:41
Solve 𝝏𝒖/𝝏𝒕=(𝝏^𝟐 𝒖)/𝝏𝒙^𝟐 (𝟎, 𝟏) at t=0.002 𝒖=(𝟎,𝒕)=𝟎=𝒖(𝟏,𝒕), 𝒖(𝒙,𝟎)={(πŸπ’™ (𝟎, 𝟎.πŸ“) @𝟐(πŸβˆ’π’™), (𝟎.πŸ“, 𝟏)
02:41
Find the numerical solution (𝝏^𝟐 𝒖)/𝝏𝒙^𝟐 =𝟐 𝝏𝒖/𝝏𝒕 , 𝒖=(𝟎,𝒕)=𝟎=𝒖(πŸ’,𝒕), 𝒖(𝒙,𝟎)=𝒙(πŸ’βˆ’π’™) ,h=1,t=4 steps.
02:01
Numerical Solution of the one dimensional Heat equation & working procedure.
02:01
Show that: π’Š)𝑷_𝟐(π’„π’π’”πœ½)=1/4 (𝟏+πŸ‘π’„π’π’”πœ½), π’Šπ’Š) 𝑷_πŸ‘ (π’„π’π’”πœ½)=𝟏/πŸ– (πŸ‘π’„π’π’”πœ½+πŸ“π’„π’π’”πŸ‘πœ½).
02:01
Express 𝒇(𝒙)=𝒙^πŸ’βˆ’πŸ‘π’™^πŸ‘βˆ’π’™^𝟐+πŸ“π’™ in terms of Legendre polynomial.
02:01
Express πŸπ’™^πŸ‘βˆ’π’™^πŸβˆ’πŸ‘π’™+𝟐 in terms of Legendre polynomial
02:01
If 𝒙^πŸ‘+πŸπ’™^πŸβˆ’π’™+𝟏=𝒂𝑷_𝟎 (𝒙)+𝐛_𝑷 𝟏( 𝒙)+c𝑷 _𝟐 (𝒙) +d𝑷_πŸ‘ (𝒙) .Find the value of a, b, c, d.
01:21
Express 𝒙^πŸ‘+πŸπ’™^πŸβˆ’πŸ’π’™+πŸ“ in terms of Legendre polynomial
02:01
Rodrigues Formula & Results.
02:01
Series Solution of Legendre s Differential Equation
04:01
Orthogonality of Bessel s Functions
02:01
Prove that : 𝑱((βˆ’πŸ)β„πŸ) (𝒙)=√(𝟐/𝝅𝒙) 𝒄𝒐𝒔𝒙.
02:41
Prove that : 𝑱(πŸβ„πŸ) (𝒙)=√(𝟐/𝝅𝒙) π’”π’Šπ’π’™.
05:22
Bessel's Functions: Series solution of Bessel's Differential Equation Leading to Bessel's Function.
01:21
Solve the wave equation (𝝏^𝟐 𝒖)/〖𝝏𝒙〗^𝟐 =(𝝏^𝟐 𝒖)/〖𝝏𝒕〗^𝟐 𝒖(𝒙,𝟎)=π’”π’Šπ’π…π’™ (0,1), h=0.2, k=0.2 at 𝒕=𝟏.
02:01
Solve the wave equation 25 (𝝏^𝟐 𝒖)/𝝏𝒙^𝟐 =(𝝏^𝟐 𝒖)/𝝏𝒕^𝟐 𝒖(𝒙,𝟎)={(πŸπŸŽπ’™ πŸŽβ‰€π’™β‰€πŸ, πŸ“(πŸ“βˆ’π’™) πŸβ‰€π’™β‰€πŸ“) h=1, πŸŽβ‰€π’•β‰€πŸ.
02:41
Solve the Wave equation (𝝏^𝟐 𝒖)/(𝝏𝒙^𝟐) =𝟎.πŸŽπŸ”πŸ“ (𝝏^𝟐 𝒖)/𝝏𝒕^𝟐 , 𝒖(𝒙,𝟎)=𝒙^𝟐 (π’™βˆ’πŸ“), h=1, k=0.5 ,4 steps.
02:21
Solve the wave equation (𝝏^𝟐 𝒖)/〖𝝏𝒕〗^𝟐 =πŸ’ (𝝏^𝟐 𝒖)/〖𝝏𝒙〗^𝟐 𝒖(𝒙,𝟎)=𝒙(πŸ’βˆ’π’™) , h=1, k=0.5 upto 4 steps.
01:31
Working procedure for one dimensional wave equation
02:31
21MAT31: Numerical solution of the one dimensional wave equation.
02:04
21MAT31:Numerical Method PDE, Classification of PDE of second order & Finite difference approximate.
06:48
# Partial Derivatives || # Definition || #Symbols
06:43
#Evaluate || lim_π’™β†’πŸŽβ‘(π’”π’Šπ’π’™π’™^πŸβ„π’™^𝟐 ) || 21MAT11 || Module : 2 || Differential Calculus ||BMATE101
06:32
Evaluate || lim_π’™β†’πŸŽβ‘(𝒕𝒂𝒏𝒙𝒙^πŸβ„π’™ ) || 21MAT11 || Module : 2 || Differential Calculus ||
05:58
Evaluate || lim┬(π’™β†’πŸŽ)⁑[((𝒂^𝒙+𝒃^𝒙+𝒄^𝒙+𝒅^𝒙 ))/πŸ’]^(πŸβ„π’™) || 18MAT11 || Differential Calculus
05:15
Evaluate || lim_π‘₯β†’πœ‹/2⁑(π’”π’Šπ’π’™^𝒕𝒂𝒏𝒙 )|| 18MAT11 || Differential Calculus
03:58
Evaluate || lim┬(π’™β†’πŸŽ)⁑ ([𝒄𝒐𝒔𝒙]^(πŸβ„π’™^𝟐 )) || Differential Calculus || #Math E Connect
04:38
Evaluate || lim_(π’™β†’πŸ)⁑(𝒙^(πŸβ„(πŸβˆ’π’™)) )|| Differential Calculus || #Math E Connect||
04:16
How to evaluate (𝟎)^𝟎 , (∞)^𝟎 (𝟏)^∞ || Differential Calculus. ||
11:43
Evaluate || lim_(π’™β†’πŸŽ)⁑[𝟏/𝒙^𝟐 βˆ’πŸ/((π’”π’Šπ’)^𝟐 𝒙)] || Differential Calculus ||
12:55
Evaluate || lim_(π’™β†’πŸŽ)⁑[πŸπ’™^𝟐 βˆ’(𝒄𝒐𝒕)^𝟐 𝒙] || Differential Calculus ||
06:11
Evaluate || lim_(xβ†’0)⁑[(tanx-x)/(x^2 tanx)] ||..
04:27
Evaluate || lim_(xβ†’a) log⁑[2-x/a]cot(x-a) ||.
04:55
Evaluate || lim_(xβ†’0)⁑[1/x-(log(1+x))/x^2 ].
05:44
Evaluate the ( lim)_(xβ†’0) ⁑[(1-cosx)/(xlog(1+x))] || Differential Calculus ||
08:09
|| Evaluate || (lim⁑ )_(xβ†’1)⁑[ x/(x-1)-1/logx ] || Differential Calculus ||
05:18
Evaluate || lim π’™β†’πŸŽβ‘γ€–π’π’π’ˆπ’™ / 𝒄𝒐𝒔𝒆𝒄𝒙〗|| Differential Calculus ||
14:25
#18 Problem#4||PDE by direct integration||(𝝏^𝟐 𝒛)/ππ’™ππ’š=π’”π’Šπ’π’™π’”π’Šπ’π’š,||𝝏𝒛/ππ’š=βˆ’πŸπ’”π’Šπ’π’š, 𝒙=𝟎 and 𝒛=𝟎 π’š=𝝅/𝟐||
06:40
#17||Problem#3||Solution of non-homogenous PDE by direct integration (𝝏^𝟐 𝒛)/(𝝏𝒙^𝟐 ππ’š)=𝒄𝒐𝒔(πŸπ’™+πŸ‘π’š)||
05:30
#16 || Problem#2 ||Solution of non-homogenous PDE by direct integration Solve: (𝝏^𝟐 𝒛)/ππ’™ππ’š=𝒙/π’š+𝒂 ||
07:48
#15 || Problem#1|| Solution of non-homogenous PDE by direct integration || Solve:(𝝏^𝟐 𝒖)/𝝏𝒙^𝟐 =𝒙+π’š||
05:38
#14 ||Problem#2 ||Formation of partial differential equations|| Form a PDE from𝒇(𝒙^𝟐+π’š^𝟐, π’›βˆ’π’™π’š) ||
10:23
#13 || Problem#1 || Formation of PDE || Form the PDE fromβˆ…(𝒙+π’š+𝒛,𝒙^πŸβˆ’π’š^πŸβˆ’π’›^𝟐) || 18MAT21||
04:48
#12 || Formation of PDE by from βˆ…(𝒖,𝒗)=𝟎 where u v are functions of x, y, z. || Steps||
08:11
#30 || Solution of Lagrange’s linear Partial differential equation || Solve: (π’š^𝟐+𝒙^𝟐)𝒑+𝒙(π’šπ’’βˆ’π’›)=𝟎 ||
10:19
#29 || Solution of Lagrange’s linear Partial differential equation || Solve: π’™π’‘βˆ’π’šπ’’=π’š^πŸβˆ’π’™^𝟐|| 18MAT21
04:18
#28 ||Solution of Lagrange’s linear Partial differential equation ||Solve: π’‘π’šπ’›+𝒒𝒛𝒙=π’™π’š|| 18MAT21||
06:19
#27 || Problem on Solution of Lagrange’s linear PDE|| Solve: (π’š^𝟐 𝒛)/𝒙 𝒑+𝒙𝒛𝒒=π’š^𝟐|| 18MAT21|| #PDE||
03:21
#26 || Solution of Lagrange’s linear Partial differential equation || Working Rule ||18MAT21||#PDE||
20:02
#24 || Solution of one dimensional Heat equation (𝝏^𝟐 𝒖)/𝝏𝒕^𝟐 =𝒄^𝟐 (𝝏^𝟐 𝒖)/𝝏𝒙^𝟐 || 18MAT21 || #PDE.
19:14
#25 || Solution of one dimensional Wave equation 𝝏𝒖/𝝏𝒕=𝒄^𝟐 (𝝏^𝟐 𝒖)/〖𝝏𝒙〗^𝟐 || 18MAT21 || #PDE
06:14
#21 || Problem#7 || (𝝏^𝟐 𝒛)/〖𝝏𝒙〗^𝟐 =𝒂^𝟐 𝒛, given that when 𝒙=𝟎 and 𝒛=𝟎 and 𝝏𝒛/𝝏𝒙=π’‚π’”π’Šπ’π’š|| 18MAT21 ||
06:12
#20 || Problem#6 || (𝝏^𝟐 𝒛)/〖𝝏𝒙〗^𝟐 +𝒛=𝟎, given that when 𝒙=𝟎 and 𝒛=𝒆^π’š and 𝝏𝒛/𝝏𝒙=𝟏 ||18MAT21||#PDE
07:14
#19 || Problem#5 || (𝝏^𝟐 𝒛)/𝝏𝒙^𝟐 =π’™π’š, subject to the condition 𝝏𝒛/𝝏𝒙=π’π’π’ˆ(𝟏+π’š) when 𝒙=𝟏 and 𝒛=𝟎||
06:16
#11 || Problem#6 || PDE by eliminating the arbitrary function of the equation𝒛=π’šπ’‡(𝒙)+π’™βˆ…(π’š) ||
05:34
#10 || Problem#5 || PDE by eliminating the arbitrary function of the equation𝒛=𝒆^(𝒂𝒙+π’ƒπ’š) 𝒇(π’‚π’™βˆ’π’ƒπ’š) ||
02:47
#9 || Problem#5 || PDE by eliminating the arbitrary function of the equation 𝒛=𝒙+π’š+𝒇(π’™π’š) || 18MAT21
03:14
#8 || Problem#4 || Form a PDE by eliminating the arbitrary function of the equation𝒛=𝒆^π’š 𝒇(𝒙+π’š) ||