Derivatives of trigonometric functions

[TsAmE]

Homework Statement

Find the constants A and B such that the function y = Asinx + Bcosx satisfies the differential equation y'' + y' - 2y = sinx

Homework Equations

None

The Attempt at a Solution

My attempt: y = Asin x + Bcosx

y' = Acosx - Bsinx

y'' = - Asin x - Bcosx

y'' + y' - 2y = sin x
- Asinx - Bcosx + Acosx - Bsinx - 2Asinx + 2Bcosx = sinx
- 3Asinx + Bcosx + Acosx - Bsinx = sinx...

I don't know if my working out is correct, but I couldn't figure out what to do from there.

 

[Mr.Miyagi]

You're almost there.

Now factorize the sines and the cosines: (-3A-B)sinx + (-3B+A)cosx=sinx.

For this to be true we must have -3A-B=1 and -3B+A=0.

I think you'll be able to take over from here, right?

edit: Oops, thanks radou. Equations have been corrected

 

[radou]

"+ 2Bcosx " sold be "- 2Bcosx".

 

[TsAmE]

For this to be true we must have -3A-B=1 and -3B+A=0.

Howcome you spilt the equations into 2? and why does the one equal 1 and the other equal 0? Aren't they suppose to represent the rads of the sin and cos graph?

 

[Gavins]

You have (-3A-B)sinx + (-3B+A)cosx = 1sinx + 0cosx.

This equation has to hold for all x. So the only way this will work is if you have -3A-B = 1 and -3B+A=0.

 

[TsAmE]

Oh ok that sort of makes sense. What rule is this by the way? Cause I have never heard it when doing trig before

Edit: Nevermind I understand now :) Thanks

 

[gabbagabbahey]

Oh ok that sort of makes sense. What rule is this by the way? Cause I have never heard it when doing trig before

Whenever you have two orthogonal functions $f(x)$ and $g(x)$, and an equation of the form $\alpha f(x)+\beta g(x)=\gamma f(x)+\delta g(x)$, the only way it can be satisfied for all $x$ is if $\alpha=\gamma$ and $\beta=\delta$.This can be proved using the definition of orthogonality. Sin(x) and Cos(x) are orthogonal functions.

If you are unfamiliar with orthogonality of functions, you can also prove that $\alpha\sin x+\beta\cos x=0$ can only hold for all $x$ if $\alpha=0$ and $\beta=0$ by differentiating the equation, and solving the system of two equations (the original and the differentiated one), for your two unknowns.