Suppose that L is a linear differential operator and L(cosx) doesn't equal to 0. and L(sinx) doesn't equal to 0. Prove that the equation Ly=cosx has a solution of the form A cosx + B sinx

Question

usukidoll · Answer

@hartnn

usukidoll · Answer

so... hmm is there a complex root involved or...

usukidoll · Answer

if I just plug in cosx to Ly=cosx I get L(cosx) = cosx
for sinx to Ly=sinx  L(sinx) = sinx

usukidoll · Answer

so... L(cosx) = A 
L(sinx) =B

A = cos x
B = sin x

A+ B 
cos x + sin x? my guess

usukidoll · Answer

@wio

anonymous · Answer

Linear differential operator means it takes a function and turns it into a linear nth order differential expression, right?

anonymous · Answer

What about induction on n?

hartnn · Answer

try converting cos into its complex form, 
$\Large \cos x=\dfrac{e^{ix}+e^{-ix}}{2}$

usukidoll · Answer

:/ that looks like the hyperbolic meaning remember cosh?>

usukidoll · Answer

induction on n?

hartnn · Answer

$\Large \cos hx=\dfrac{e^{x}+e^{-x}}{2}$

anonymous · Answer

Like if $L$ is an $n$th order linear differential operator?

usukidoll · Answer

like for example let's have it at Ly would it be the ...oh no it's going to turn nasty

usukidoll · Answer

do I convert sinx into the sinhx complex form too?

anonymous · Answer

If you start with $n=0$ then you have: $$
a_1y=\cos x
$$

anonymous · Answer

Leting $A=\frac 1{a_1}$ and $B=0$ this is true.

anonymous · Answer

Then try to prove for $n$ assuming that $n-1$ and $n-2$ etc would work.

anonymous · Answer

Also, hartnn wasn't using coshx and sinhx he's using the euler definition cosx + isinx = e^ix to rewrite cosx and sinx in terms of exponentials.

anonymous · Answer

So you have $$
a_ny^{(n)}+L_2(y) = \cos x
$$Where $L_2(y)$ is some $n-1$th order linear differential operator, and we can assume by induction it has a solution in the form $A\cos x +B\sin x$

anonymous · Answer

@UsukiDoll Are you understanding my induction here?

usukidoll · Answer

oh yeah

usukidoll · Answer

I haven't gotten to Euler's version of sinx and cosx, but I still remember euler's method

usukidoll · Answer

sorry was doing my english assignment...reading my model writer's work

usukidoll · Answer

so I should do the same thing for sin x... 
an y^n+L1y = sin x?

anonymous · Answer

You only have to do it for $Ly=\cos x$ though... right?

usukidoll · Answer

oh yeah right... so no need for sin x

usukidoll · Answer

so for n-1 
An-1y^(n-1)+L2(y)=cos x