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2nd order ODE's

Hi I'm stuck and was wondering can someone explain the general method for solving them. So from the form A(x)y''+B(x)y'+C(x)y=D(x). And how does this method differ if D(x) is simply 0? And what are the partial solutions and how would the form y(x)=x^a come into it.

Also any explination of the theorem on existence and uniqueness of solutions would be gratefully recieved. :/

bump, help!

There's no general solution to differential equations. This shouldn't be too suprising because we can't find an anti-derivative for every function, i.e.

dy/dx = sin(x^2) is only a 1st order ODE and we can't even write down y as an anti-derivative here.

Things only get worse in higher dimensions.

No idea what you mean by the x^a question.

Reply 3

Josh Jones

I know theres no specific general solution unless you have initial conditions, but can you not find a general solution in terms of constants c1 and c2 or whatever?

The example I'm considering is:

a) Find a general solution of (x^2)y''-4xy'+6y=0
Hint: Seek for the partial solutions in the form y(x)=x^a where a is a constant to be determined.

b) Shot the initial value problem y(0)=yo, y'(0)=y1 for that equation has a solution only if yo=y1=0. Does it mean that the theorem on existence and uniqueness of solutions is wrong? Explain your answer.

So what are the things to be thinking about when solving this 2nd order ODE? And more generally what different forms are there that lead to different types of solutions?

Reply 4

boredom_personified

As has already been pointed out, there is no general solution to the general 2nd order differential equation you posted earlier, and it would largely depend upon what A(x), B(x) and C(x) as to whether such a solution even exists without finding it as a series solution.
I'm not entirely sure what you mean with your x^a question, but i'm suspecting what it means is that x^a is an eigenfunction of the differential operator

x \cdot \frac{d}{dx}

in the same way that

e^{\lambda x}

is an eigenfunction of

\frac{d}{dx}

. This will work in your specific example because in your example, the nth derivative of y has an

x^n

expression as it's nonconstant coefficient (when you have

y=x^a

then

y'=ax^{a-1}

and

y'' = a(a-1)x^{a-2}

.This means that in your example, where they are multiplied by the correct coefficient, you can take

x^a

out as a factor from the original equation to get your 'auxillary' equation):

y=x^a

y'=ax^{a-1}

and

y'' = a(a-1)x^{a-2}

.
Substituting into

(x^2)y''-4xy'+6y=0

a(a-1)x^2 \cdot x^{a-2} -4ax \cdot x^{a-1} + 6x^a = 0

then using indices laws to neaten this up

\Longrightarrow a(a-1)x^a -4ax^a + 6x^a = 0 \Longrightarrow a(a-1) -4a +6 = 0

allowing you to solve for

a

and find a solution to your equation.

Reply 5

Josh Jones

ok thankyou i understand this example now. last point is what are the implications that for this equation only has a solution when yo=y1=0 on the theorem of existence and uniqueness of solutions? am i correct in thinking that the theorem states for all yo and y1 there is a unique solution that exists whereas in this example it only has a solution when the initial conditions are 0?

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2nd order ODE's

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