How did this second-order initial value problem do this?

[myusernameis]

Homework Statement

Find a second-order initial-value problem whose solution is sin at

The Attempt at a Solution

so i know that

y' = a cos at
y" = -a^2 sin at

but the solution says that

"Then y(t) = sin at is
a solution to 2 y ''+ a y = 0 . Our initial conditions must be y(0) = 0, y '(0) = a ."

how do i know that it's 2y'' + a y? how come it couldn't be something else like y''+y'+ y?

thanks

 

[djeitnstine]

This is because y"+y=0 is the only DE that has the solution of the form Acos kt + Bsin kt, other DE's have a solution of the form $$Ae^{bt}+...$$ such as the one you posted. (y"+y'+y). This stems from the characteristic equation $$r^{2}+r=0$$ which has complex roots...

 

[gabbagabbahey]

"Then y(t) = sin at is
a solution to 2 y ''+ a y = 0 . Our initial conditions must be y(0) = 0, y '(0) = a ."

That can't be right; $y(t)=\sin(at)$ is not the solution to that DE, $y(t)=\sqrt{2a}\sin(\sqrt{\frac{a}{2}}t)$ is.

Either you've made a typo/error, or the solution manual has a typo/error or some combination of the two.

how come it couldn't be something else like y''+y'+ y?

thanks

Well, as you've found, if $y(t)=\sin(at)$, then $y'(t)=a\cos(at)$ which is orthogonal to both y(t) and y''(t) (you can't add some non-zero multiple of cos(at) to some polynomial of sin(at) and get a constant or zero), so the DE can't contain a y'(t) term...

 

[HallsofIvy]

Yes, if y= sin(at), then y'= -a cos(at) and y"= -a²sin(at).

The simplest second order differential equation y= sin(at) satisfies is y"+ a²y= 0. I don't know where you got that "2".

(I wouldn't be surprised if this were the answer to "problem 2"!)

 

[myusernameis]

Yes, if y= sin(at), then y'= -a cos(at) and y"= -a²sin(at).

The simplest second order differential equation y= sin(at) satisfies is y"+ a²y= 0. I don't know where you got that "2".

(I wouldn't be surprised if this were the answer to "problem 2"!)

thanks!

turns out there wasn't a "2" anywhere...