Vector of differential solutions

[boneill3]

Homework Statement

Hi Guy's

Let V be the set of functions $f:R\rightarrowR$which solve the differential equation:

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

Show that $e_1:R\rightarrow R, x \rightarrow e^x$ and $e_2:R\rightarrow R, x\rightarrow cosh(x)$
Comprise a basis for V.

Homework Equations

The Attempt at a Solution

I have not done 2nd order differential equations and I was wondering if some one might give me a hint to a particular solution vector.

I know I need to prove that $e^x$ and $cosh(x)$ are linear independent to be a basis, but I need to show that any solution vector can be generated by them.

with the natural log we can't take the log of zero so when can the vector equal zero?

 

[HallsofIvy]

Well, first of all, your $e_1$ and $e_2$ won't comprise a basis for the space of all functions- that's infinite dimensional. What you mean is that V is the set of all functions that satisfy that differential equation.

That means you need to show the properties of a basis:
1) They are independent. In this case that means that the only way you can have $Ae^x+ Bcosh(x)= 0$, for all x, is if A= B= 0.

2) They span the space. That is, that if y is any function satifying the differential equation, then $y= Ce^x+ Dcosh(x)$ for some numbers C and D.

 

[boneill3]

2) They span the space. That is, that if y is any function satifying the differential equation, then $y= Ce^x+ Dcosh(x)$ for some numbers C and D.

Thankyou for your explanation.
I don't know how I am going to prove this last statement. Do I need to have an explicit solution vector to show that it can be generated by $y= Ce^x+ Dcosh(x)$ ?

 

[boneill3]

when you are dealing with these 2nd order differential equations does it mean that there are two solutions?

So if I prove that

$y= Ce^x+ Dcosh(x)= 0$ if and only if the constants C and D are zero than that basis can generate the solution set of vectors V.

 

[boneill3]

As the functions inverses $e^x$ and $cosh(x)$
are not defined for 0 ie ln(0) is undifined etc is that enough to say that they must be independent as they can only equal zero if both constants are zero?

 

[HallsofIvy]

No, whether they are indpendent or not has nothing to do with inverses.

Suppose Ae^x+ Bcosh(x)= 0 for all x. In particular, Ae⁰+ B cosh(0)= A+ B= 0. Further, if Ae^x+ Bcosh(x)= 0, a constant, for all x, then its derivative must be 0 for all x also: Ae^x+ Bsinh(x)= 0 for all x. Now take x= 0.

To show that these span the set of all solutions to this second order equation, suppose y is such a solution. Let A= y'(0) and B= y(0)- y'(0). Then, as before, z(x)= Ae^x+ B cosh(x) is a solution to the same differential equation (of course, you have shown that e^x and cosh(x) satisfy the differential equation?) having z(0)= A+ B= y(0) and z'(0)= A= y'(0). Now use the "existence and uniqueness theorem".

 

[tiny-tim]

Hi boneill3!

(try using the X² tag just above the Reply box )

Hint: put z = dy/dx + y

 

[boneill3]

Isn't the "existence and uniqueness theorem". mean that it is an solution in the space if the function and the first paritial derivative are continuous .

You say Hint: put z = dy/dx + y

we have
$\frac{d^2y}{dx^2}=y$
so

$\frac{dy}{dx}=yx$

so

$z = yx+y$
and

$z' = y$

which are both continuous proving that Z is in the solution space

 

[tiny-tim]

$\frac{d^2y}{dx^2}=y$
so

$\frac{dy}{dx}=yx$

nooo … you can't integrate y and get yx

Differentiate z = dy/dx + y …

what do you get? ​

 

[boneill3]

how do you differentiate dy/x ? what is its value?

 

[HallsofIvy]

He didn't say "dy/x" he said "dy/dx". And its derivative if the second derivative, $d^2y/dx^2$. The derivative of z= dy/dx+ y is $d^2y/dx^2+ dy/dx$.

But I must admit, tiny tim, I don't see how that is related to this problem!