Differentiating y= Acosαx + Bsinαx and finding Constants

[zombieguy]

Homework Statement

Don't really know what to do with this one, any help welcome

If A, B and α are constants, independent of x, and

y= Acosαx + Bsinαx

Evaluate the derivates
y'=dy/dx
y^n=d^2y/dx^2

Determine values of A, B and α such that y satisfies
y^n+y=0 and y(0)=0, y'(0)=1

Homework Equations

The Attempt at a Solution

I've tried to differentiate y
y'=-αAsinαx + αBcosαx
y'=αBcosαx - αAsinαx

y^n=-ααBsinαx + ααAcosαx
y^n=ααAcosαx - ααBsinαx

 

[LCKurtz]

You might find it easier to see if you used standard notation and parentheses. Use y'' for second derivative and use exponents like a² instead of aa. (Use the X² icon for superscripts.

Rewriting what you have for the second derivative:

y'' = a²Acos(αx) - α²Bsin(αx)

Now can you figure out a so that

y''+y=0

And once you have that what do the two conditions

y(0)=0, y'(0)=1

tell you about A and B?

 

[zombieguy]

But are my y' and y'' values correct because their wouldn't be any point in starting part 2 of the question if they are not

 

[LCKurtz]

You have a sign mistake in y'' which will become obvious when you try to make y'' + y = 0.

 

[zombieguy]

y''=-ααBsinαx - ααAcosαx then, yes?

 

[LCKurtz]

Yes. Did you understand about the X² button so you can write a² instead of aa?

 

[zombieguy]

y+yⁿ=0

-α²Bsinαx - α²Acosαx + Acosαx + Bsinαx=0

Acosαx + Bsinαx=α²Bsinαx + α²Acosαx

Acosαx(1 - α²)=Bsinαx(α² - 1)

so α=1?

b) 0=Acosα(0) + Bsinα(0)
A=0

c) 1=αBcosα(0) - αAsinα(0)
B=1

 

[LCKurtz]

y+yⁿ=0

-α²Bsinαx - α²Acosαx + Acosαx + Bsinαx=0

Right here, if you collect terms on cos and sin you can see what a has to be.

Acosαx + Bsinαx=α²Bsinαx + α²Acosαx
Divide by cosαx
A + Btanαx=α²Btanαx + α²A

A (1 - α²)=Btanαx(α² - 1)

so α=1?

Is a = 1 the only solution to a²=1?

b) 0=Acosα(0) + Bsinα(0)
A=0

c) 1=αBcosα(0) - αAsinα(0)
B=1

That's right if a = 1 is the only value for a that works.

 

[zombieguy]

Is a = 1 the only solution to a²=1?

My bad, a=1 or -1
B=1 or -1
A=0

Thank you very much for the help