Differential equations (1 Viewer)

[hscishard]

How do you solve them?

E.g something like y''+2y'+y=0

 

[Trebla]

How do you solve them?

E.g something like y''+2'+y=0

What do you mean by 2'?

It depends on the nature of the differential equation (e.g. whether it is linear, separable etc). Solving second order differential equations are currently not in the course.

 

[darkchild69]

Isn't the general solution for second order differential equation:

y = Ae^2x + Be^x

but don't you require boundary conditions to solve specific equations?

what is 2'?

 

[hscishard]

Whoops 2y' sorry

 

[hscishard]

What do you mean by 2'?

It depends on the nature of the differential equation (e.g. whether it is linear, separable etc). Solving second order differential equations are currently not in the course.

I swear I saw that somewhere in the Cambridge book.

 

[DNETTZ]

Dont worry it is in the Cambridge book, in Ch. 13 somewhere.

y''+2y'+y=0

You have to use the other compoments of the question to solve this, btw. Exp. function is good for that.

 

[Entrant]

you get y, y', and y" and substitute them into the equation. And BTW its in the Trig Part of CAMB 3U YR11.
e.g.
y=x^3
therefore.
y' = 3x^2
and y" = 6x
therefore substituting into your equation
6x-6x^2+x^3 = blah blah.
I know its not equal to 0 but its an example on how to substitute it back in.

 

[hscishard]

So I guess you can't solve it whenever you're not given anything? Like no y or y',y'',etc.

 

[Entrant]

yeah.
they have you give you a y.

 

[tommykins]

So I guess you can't solve it whenever you're not given anything? Like no y or y',y'',etc.

You normally guess y = e^(ax) for some a and solve for a.

 

[DNETTZ]

You normally guess y = e^(ax) for some a and solve for a.

What?
Then you'd get something strange with 2 variables in a and x, and that cant be solved with a single eqn as I recall?

a^2*e^ax+2(ae^ax)+e^ax=0?
Unsolvable without 'a' value? so you cant really solve for y?

You need a f(x) to solve any diff eqn, and it doesnt necessarily have to be exponential f(x)

 

[tommykins]

What?
Then you'd get something strange with 2 variables in a and x, and that cant be solved with a single eqn as I recall?

a^2*e^ax+2(ae^ax)+e^ax=0?
Unsolvable without 'a' value? so you cant really solve for y?

You need a f(x) to solve any diff eqn, and it doesnt necessarily have to be exponential f(x)

Lol?

e^ax(a^2+2a+1) = 0

e^ax =/= 0 so thus a^2+2a+1 = 0.

And lol, of course you need a simple guess, you solve the homogenous equation and if the RHS of the ODE is a function g(x) then you guess the most general solution to g(x).

If you haven't done ODE's before then I'd advise you to be quiet.

 

[Trebla]

What?
Then you'd get something strange with 2 variables in a and x, and that cant be solved with a single eqn as I recall?

a^2*e^ax+2(ae^ax)+e^ax=0?
Unsolvable without 'a' value? so you cant really solve for y?

You need a f(x) to solve any diff eqn, and it doesnt necessarily have to be exponential f(x)

e^ax cancels off as it is non-zero which leaves an equation in terms of a only. You then have what's called the characteristic polynomial to solve and the solutions you get for a (say a₁ and a₂ assuming they are distinct) allow the general solution to be y = Ae^a₁x + Be^a₂x where A and B are arbitary constants. The reason the general solution takes that form is because of something known as the law of superposition which holds for linear differential equations.

 

[DNETTZ]

Lol?

e^ax(a^2+2a+1) = 0

e^ax =/= 0 so thus a^2+2a+1 = 0.

ah, i didnt think of that. Nice. I'm useless without a piece of paper
:uhhuh::uhhuh::uhhuh::uhhuh:

 

[tommykins]

ah, i didnt think of that. Nice. I'm useless without a piece of paper
:uhhuh::uhhuh::uhhuh::uhhuh:

Trebla can correct me, but as far as linear second order ODE's of the form ay'' + by' + cy = d for some a,b,c,d - you always 'guess' y = e^zx for some z.

 

[DNETTZ]

I'm going to go and do some implicit differentiation now, because I just realised that I have forgotten it (OH THE HORRRRRORS IN STORE FOR ME!)

 

[ferdin]

aux eqn is m^2+2m+1 =0
( m+1)^2=0
m=-1,-1
roots are equal
solution is:
y= (A+BX)e^(-x)