Room for Yuvraj and robjohn

Jack Rod

5:43 AM

@robjohn hi

sir can u help me in understanding the derivation of linear differentiation equation of nth order

let i assume differential equatio to be ...

→.

$n\frac {dy^n}{dx}+n-1\frac{dy^{n-1}}{dx}....................\frac{dy}{dx}=0$

how do I evaluate it

what I did I assumed a standard solution of the equation to be y=cx

so solution would be $c_nx+c_{n-1}x..........................cx=0$

now I assumed a simple linear differential equation of second degree that

is 3y''+4y'-3y=0

there comes out be two roots

pf the equation

but they have represented these roots in the form of ce^x

IDK how ce^x comes from?

Jack Rod

6:46 AM

indirectly saying i want to understand proof of all the identity given below in the link

math24.net/….

robjohn

7:11 AM

@JackRod: Do you understand that that function satisfies the differential equation?

The characteristic equation is $3x^2+4x-3=0$, which has solutions $\alpha=\frac{-2+\sqrt{13}}{3}$ and $\beta=\frac{-2-\sqrt{13}}{3}$

If we set $y=e^{\alpha x}$, then we get $y'=\alpha y$ and $y''=\alpha^2y$

$3y''+4y'-3y=\left(3\alpha^2+4\alpha-3\right)y=0$

If we set $y=e^{\beta x}$, then we get $y'=\beta y$ and $y''=\beta^2y$

$3y''+4y'-3y=\left(3\beta^2+4\beta-3\right)y=0$

Jack Rod

yes?

sir why do we add the two roots like if i assume as above the solution of it will be

$y=c1e^{\alpha x}+c2e^{\beta x}$

robjohn

If $3y_1''+4y_1'-3y_1=0$ and $3y_2''+4y_2'-3y_2=0$, what would $3(y_1+y_2)''+4(y_1+y_2)'-3(y_1+y_2)$ be?

Jack Rod

7:29 AM

=$\left(3\beta^2+4\beta-3\right)y+\left(3\alpha^2+4\alpha-3\right)y$

robjohn

however, I didn't say what $y_1$ or $y_2$ were. I just said they satisfied that equation.

Jack Rod

yes sir

it is like ()r1+()r2?

where r1 and r2 are the roots?

robjohn

$(y_1+y_2)'=y_1'+y_2'$ right?

Jack Rod

yes

robjohn

then $(y_1+y_2)''=y_1''+y_2''$ right?

Jack Rod

7:32 AM

yess

robjohn

If $3y_1''+4y_1'-3y_1=0$ and $3y_2''+4y_2'-3y_2=0$, what would $3(y_1+y_2)''+4(y_1+y_2)'-3(y_1+y_2)$ be?

Jack Rod

3y1"+3y2"+4y1'+4y2'-3y1-3y2

and 3y"1+4y1'-3y1+3y"2+4y'2+-3y2=

@robjohn hi

robjohn

and what does that equal?

Jack Rod

y

that is a general solution

11 mins ago, by Jack Rod

it is like ()r1+()r2?

robjohn

$3(y_1+y_2)''+4(y_1+y_2)'-3(y_1+y_2)=\left(3y_1''+4y_1'-3y_1\right)+\left(3y_2''+4y_2'-3y_2\right)=0+0=0$

Jack Rod

7:43 AM

oh shit yes

they are roots

yes yes zero

robjohn

so if you add any two solutions to that equation, you get another solution.

Jack Rod

sir if roots are equal then?

robjohn

if $3y''+4y'-3y=0$ what is $3(ay)''+4(ay)'-3(ay)$?

for a constant $a$

Jack Rod

$(3(y)''+4(y)'-3(y))a$

robjohn

and ?

Jack Rod

7:47 AM

=0

robjohn

yes

so if $y_1$ and $y_2$ are solutions, and $a_1$ and $a_2$ are constants, then $a_1y_1+a_2y_2$ is also a solution.

Jack Rod

                                    ok sir

robjohn

That is true of any homogeneous linear differential equation

Jack Rod

sir like the question above roots are in complex form

37 mins ago, by robjohn

The characteristic equation is $3x^2+4x-3=0$, which has solutions $\alpha=\frac{-2+\sqrt{13}}{3}$ and $\beta=\frac{-2-\sqrt{13}}{3}$

in general if iassume $y1=e^{\alpha x}$

y=y1+y2?@robjohn

robjohn

$a_1e^{\alpha x}+a_2e^{\beta x}$ satisfies $3y''+4y'-3y=0$

Jack Rod

7:58 AM

sir if I did not add a1 and a2 then will it not be the solution?

robjohn

@JackRod what, you mean set $a_1=a_2=1$?

Jack Rod

yes

robjohn

that would be a solution, but not all solutions.

Jack Rod

24 secs ago, by robjohn

that would be a solution, but not all solutions.

sorry sir but what does it mean

robjohn

$e^{\alpha x}+e^{\beta x}$ is a solution to the equation but so is $2e^{\alpha x}+e^{\beta x}$

$3e^{\alpha x}+5e^{\beta x}$ would be another solution

Jack Rod

8:06 AM

one differential equation can satisfy many equations?

robjohn

No. The differential equation is $3y''+4y'-3y=0$

$y=a_1e^{\alpha x}+a_2e^{\beta x}$ are the solutions for the $\alpha$ and $\beta$ we computed above.

Jack Rod

ok sir I have one mre simple example if u can explain through that

robjohn

$a_1$ and $a_2$ are chosen to satisfy initial conditions (or something similar)

Jack Rod

like if $\frac {d^2 y}{dx}-y=0$

robjohn

$\frac {d^2 y}{dx^2}-y=0$

Jack Rod

8:14 AM

yes

robjohn

what are the roots of $x^2-1=0$?

why?

Jack Rod

57 secs ago, by robjohn

what are the roots of $x^2-1=0$?

=+-1

robjohn

so the solutions would be of the form $y=a_1e^x+a_2e^{-x}$

Jack Rod

what I was saying y" i assume to be y^2

y^2-y=0

robjohn

I don't understand. $y''$ is not $y^2$. you will only confuse things if you write that.

Jack Rod

8:18 AM

ok..?

robjohn

@JackRod that has solutions $y=1$ and $y=0$, those are not related to the differential equation.

Jack Rod

yes

that is what i got

robjohn

Why are you converting $y''$ to $y^2$?

Jack Rod

because my book usually proceed with it

robjohn

Are you sure? it says to replace $y''$ with $y^2$?

that will in NO WAY lead to a solution

The characteristic equation for $y''-y=0$ is $x^2-1=0$

check that $e^x$ and $e^{-x}$ are both solutions to $y''-y=0$

Jack Rod

8:23 AM

yes they are

robjohn

that is because $+1$ and $-1$ are both roots of $x^2-1=0$

Jack Rod

ok sir so how do i get characteristic equation from the differential equation like this you have given me the equation and I just plug it and got right

but in general when i for any linear differential equation of some order n

how do i proceed with first step?

robjohn

The characteristic equation of the homogeneous linear differential equation $ay''+by'+cy=0$ is $ax^2+bx+c=0$

Jack Rod

ok

robjohn

the roots of that equation $\alpha_1$ and $\alpha_2$ will then lead to the solutions of the differential equation of the form $y=c_1e^{\alpha_1x}+c_2e^{\alpha_2x}$

Jack Rod

8:27 AM

ok sir now i got it

robjohn

it is pretty easy to see how to extend that to higher order equations, is it not?

Jack Rod

yes

sir they are two three terms in the book which i need to understand

first is :linear independence

second is the wronskian

robjohn

a set of vectors $\{v_k\}$ are linearly independent if $\sum_ka_kv_k=0$ means that $a_k=0$ for all $k$

that means you cannot write a linear combination that vanishes, unless the coefficients are all $0$

Jack Rod

ok

sir

linear independence of a linear differential equation is all the constants are zero

if the sum of it to be zero

what about second?

robjohn

no. independence is not something that is assigned to differential equation

Jack Rod

8:35 AM

@robjohn

ok

robjohn

a set of solutions can be linearly independent or not

the wronskian is a way of determining if a set of solutions are independent. Sometimes a set of independent solutions has a vanishing wronskian, but if the wronskian doesn't vanish, the solutions are independent.

Jack Rod

ok?

robjohn

Wronskian

In mathematics, the Wronskian (or Wrońskian) is a determinant introduced by Józef Hoene-Wroński (1812) and named by Thomas Muir (1882, Chapter XVIII). It is used in the study of differential equations, where it can sometimes show linear independence in a set of solutions. == Definition == The Wronskian of two differentiable functions f and g is W(f, g) = f g′ – g f ′. More generally, for n real- or complex-valued functions f1, . . . , fn, which are n – 1 times differentiable on an interval I, the Wronskian W(f1, . . . , fn) as a function on I is defined by W (...

Jack Rod

ok sir

let me have a read

@robjohn now I will practice some question and get back to u if get any trouble

robjohn

ok

Jack Rod

8:56 AM

@robjohn sir if a equation has four roots and all are equal

robjohn

10:18 AM

equal roots add a degree to a polynomial coefficient: $y''-2y'+y=0$ has solutions of $y=(a_1+a_2x)e^x$

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♦ Room for Yuvraj and robjohn