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

1. Homework Statement

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


3. 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

Gold Member
614
0
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 [tex]Ae^{bt}+...[/tex] such as the one you posted. (y"+y'+y). This stems from the characteristic equation [tex]r^{2}+r=0[/tex] which has complex roots...
 

gabbagabbahey

Homework Helper
Gold Member
5,001
6
"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; [itex]y(t)=\sin(at)[/itex] is not the solution to that DE, [itex]y(t)=\sqrt{2a}\sin(\sqrt{\frac{a}{2}}t)
[/itex] 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 [itex]y(t)=\sin(at)[/itex], then [itex]y'(t)=a\cos(at)[/itex] 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 cant contain a y'(t) term....
 

HallsofIvy

Science Advisor
Homework Helper
41,683
865
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"!)
 
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...
 

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