# Differentiating y= Acosαx + Bsinαx and finding Constants

• zombieguy
In summary, the conversation discusses determining values for constants A, B, and α in a given equation and evaluating the derivatives of the equation. The conversation also covers the conditions that must be satisfied by the equation and how to find the values for A, B, and α. Ultimately, the conversation concludes that the only solutions for the constants are A=0, B=1 or -1, and α=1 or -1.
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

## 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

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?

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

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

y''=-ααBsinαx - ααAcosαx then, yes?

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

y+yn=0

2Bsinαx - 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

Last edited:
zombieguy said:
y+yn=0

2Bsinαx - 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.

LCKurtz said:
Is a = 1 the only solution to a2=1?

B=1 or -1
A=0

Thank you very much for the help

## What is the purpose of differentiating y= Acosαx + Bsinαx?

The purpose of differentiating this equation is to find the rate of change of the function at any given point. This can help determine the slope of the curve or the velocity of an object in motion.

## How do you differentiate y= Acosαx + Bsinαx?

To differentiate this equation, you can use the trigonometric identities for the derivatives of cosine and sine. The derivative of cosine is -Asinαx and the derivative of sine is Bcosαx. You can also use the power rule to differentiate the αx term.

## What are the constants A and B in the equation y= Acosαx + Bsinαx?

The constants A and B represent the amplitudes of the cosine and sine functions, respectively. They determine the height of the curve and affect the period and frequency of the function.

## How do you find the value of A and B in the equation y= Acosαx + Bsinαx?

To find the values of A and B, you can use the initial conditions of the function. This means plugging in the coordinates of a known point on the curve into the equation and solving for A and B. You can also use the period or frequency of the function to find the values of A and B.

## What are some real-life applications of differentiating y= Acosαx + Bsinαx?

This type of differentiation is commonly used in physics and engineering to analyze the motion of objects in oscillatory systems. It can also be used in signal processing and sound engineering to analyze sound waves. Additionally, this type of differentiation is useful in calculating the rate of change in various natural phenomena, such as the growth of populations or the spread of diseases.

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