🔵18 - Second Order Linear Homogeneous Differential Equations with Constants coefficients
In this video, we shall learn how to solve Second order linear homogeneous differential equations.
A second order linear homogeneous diff equation is given of the form
p(x)d^2y/dx^2 + q(x)dy/dx +r(x)y = 0, where p, q and r are all functions of x.
To solve such equations with constants coefficients, we simply derive the auxiliary equation or the characteristic equation of the differential equation.
This auxiliary equation is quadratic and hence its roots can be found by the
1. method of factorization or
2. quadratic formula.
Three cases may arise
if roots are;
Case 1
real and unequal / distinct
Case 2
real and equal
Case 3
complex
00:00 - Introduction
07:50 - Nature of roots
11:13 - Real and unequal / distinct roots
13:40 - IVP: Real and unequal / distinct roots
20:45 - Real and equal / roots
22:26 - complex roots
Playlists on various Course
1. Applied Electricity
• APPLIED ELECTRICITY
2. Linear Algebra / Math 151
• LINEAR ALGEBRA
3. Basic Mechanics
• BASIC MECHANICS / STATICS
4. Calculus with Analysis / Calculus 1 / Math 152
• CALCULUS WITH ANALYSIS...
5. Differential Equations / Math 251
• DIFFERENTIAL EQUATIONS
6. Electric Circuit Theory / Circuit Design
• ELECTRIC CIRCUIT THEOR...
Make sure to watch till the end.
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Thank you.
Пікірлер: 25
I enjoy your videos a lot 🇿🇲🇿🇲
@SkanCityAcademy_SirJohn
10 ай бұрын
Aww that's nice
Very nice, thank you sir
@SkanCityAcademy_SirJohn
Жыл бұрын
you are most welcome
Thank you Sir.
@SkanCityAcademy_SirJohn
Жыл бұрын
You are most welcome
pls I'm looking for higher order differential equations do you have it on your playlist?
@SkanCityAcademy_SirJohn
10 ай бұрын
It's up to 2nd order DE
Very slow, calm and clear. ❤
@SkanCityAcademy_SirJohn
Жыл бұрын
That's great
@AFCOE
Жыл бұрын
@@SkanCityAcademy_SirJohn Good morning sir. Please sir am stock to solve. y"-2y'+y = ((3x^2)+1)e^x Am stuck on the particular integral. Thanks
@AFCOE
Жыл бұрын
Am letting Yp = (ax^2+bx+c)e^x. And it is giving. a=1/2, b(disappeared), c(disappeared). Please help me out❤
@SkanCityAcademy_SirJohn
Жыл бұрын
Okay, so once ax2.... Fails, you then try ax3... In that other until it matches up
@AFCOE
Жыл бұрын
Okay sir thanks very much. So there is no way to know what level to go..? ax^n at once?
You are interesting me and also I need to know how to solve first order differential equations of higher degree
@carultch
7 ай бұрын
It's still the same fundamental idea. You just have a characteristic cubic (or higher degree polynomial), instead of a characteristic quadratic to solve. The idea is still to solve the equation assuming it's homogeneous, by assuming a solution of e^(r*t), and differentiating to match the original diffEQ. Then we solve for r, and construct an arbitrary linear combination of e^(r*t) based on all possible values of r. Similar procedures also apply for solving non-homogeneous diffEQs of higher order, by finding the particular solution and combining it with the homogeneous part.
@SkanCityAcademy_SirJohn
7 ай бұрын
Nice contribution.
Atleast i've gain something🥰
@SkanCityAcademy_SirJohn
17 күн бұрын
Thank you so much
You say "linear, because the coefficients are functions of x". But that is not the reason for the linearity of the DE. Even if the coefficients were not functions of x this equation would be a LDE. It is linear because the equation is a linear polynomial in the unknown function and its derivatives, i.e., the unknown function and all its derivative terms only appear to the maximum power of 1 (see also here en.wikipedia.org/wiki/Linear_differential_equation).
@SkanCityAcademy_SirJohn
8 ай бұрын
Thanks for your contribution, that is a great enlightenment. Thanks so much. Much love❤️
It's waaaa
@SkanCityAcademy_SirJohn
3 ай бұрын
Thanks so much bro