Question

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

Answer 1

it looks like the last option, though i may be wrong

Answer 2

Thanks, I don't understand how to find solutions to differential equations.

Answer 3

y'' of sinx is -sinx so you are back to 0 if you start with 0

Answer 4

you can try differentiating all the functions there

Answer 5

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

Answer 6

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.

Answer 7

Thanks I'll go do that now

Answer 8

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

Answer 9

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.

Answer 10

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