## anonymous one year ago 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

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1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

you can try differentiating all the functions there

5. 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

6. 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.

7. anonymous

Thanks I'll go do that now

8. 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

9. 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.

10. 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