anonymous
  • anonymous
Which of the following equations is a solution to the differential equation y"+ y=0? y = -e-x y = -ex y = cos x + sin x y = sin 2x
Calculus1
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
it looks like the last option, though i may be wrong
anonymous
  • anonymous
Thanks, I don't understand how to find solutions to differential equations.
anonymous
  • anonymous
y'' of sinx is -sinx so you are back to 0 if you start with 0

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anonymous
  • anonymous
you can try differentiating all the functions there
anonymous
  • anonymous
I think its the second last one because when you differentiate the last one you get the extra 2 from the derivative of sin(2x) so you end up with \[y''+y = -4\sin(2x) + \sin(2x) = -3\sin(2x) \not = 0\]But if you do the differentiation and substitution for \[y = \cos x + \sin x\] you'll get what you want
anonymous
  • anonymous
As for a direct solution method, if you look up how to solve "Second order linear homogeneous differential equations" that'll give you and idea.
anonymous
  • anonymous
Thanks I'll go do that now
anonymous
  • anonymous
This link is probably a good one to look at actually. http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-2ndOrderLinearEqns_Stu.pdf
IrishBoy123
  • IrishBoy123
you will find the general solution to \(y'' + y = 0\) is \(y = y_0 \ cosx + \dot y_0 \ sin x\) and in your case, \(y_o = \dot y_o = 1\) if you are unsure as to how to solve the differential equation, you would be better off plugging each suggested solution into the differential equation and seeing if it comes out at zero.
anonymous
  • anonymous
Just plug in the equations, think of it like this, in normal equations u plug in a variable to get your answer in differential equations, u plug in a function instead of a variable to get your answer for example look at \[y=-e^{-x}\] What if you plug it into \[\frac{d^2y}{dx^2}+y=0\]\[\frac{d^2}{dx^2}(-e^{-x})-e^{-x}=0\] This can also be written as and then further solved \[\frac{d}{dx}(\frac{d}{dx}(-e^{-x}))-e^{-x}=0\] If you solve this and the left side is equal to the right side, then your function satisfies the differential equation and can be considered a solution to it

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