Solved: IVP and DE Problem

[Sheneron]

[SOLVED] IVP and DE problem

Homework Statement

1) Find a solution to the IVP y" = -9y, y'(0) = 3, and y(0) = 6

2) Is y(x) = (21+6x)^1/3 a solution to the DE y' = 2/y^2

The Attempt at a Solution

For the first one I think it has something to do with sin/cos but I don't know what to do. As for the second one maybe something with y=Ae^kt. Any help would be appreciated.

 

[Sheneron]

for 2)

If i use the power rule and take the derivative, I get 2(21+6x)^-2/3. But that is so far from the question I feel like it can't be right.

 

[EngageEngage]

You have to find the characteristic equation for the first one, and find the eigenvalues. They will be imaginary and equal to +- 3i ( i think). Then you can set up a general solution

y = e^3it. Expand this with euler's formula to get sin and cos terms. Then take the derivative of y. At this point you have 2 equations and 2 ivp's, which you can solve.

 

[Sheneron]

I am in chapter 3 of calc 1, and we haven't learned about eigenvalues or Euler's formula yet. Is there any other way to do this?

 

[EngageEngage]

What kind of stuff have you been going over? the only other way that i could see doing this is by guessing. You know that the second derivative of a trig function will be the same trig function with opposite sign. if you guess y = sin(3t), y'' = -9sin(3t) which equals -9y.

 

[Sheneron]

I think that is what we were supposed to do, just guess at it. So thanks for your help there. Can you help me any with number 2?

 

[EngageEngage]

for the second problem, you just need to find the derivative of y' and plug y and y' into the given equation. If you get a true statement (0 = 0), then it is a solution.

y' = 1/3(21+6x)^(-2/3)*6 = 2/(21+6x)^(2/3) = 2/y^2. So, its a solution.

The main point of solving these problems is to go from an equation with derivatives of y to just an equation of y. To check if you have the solution, you have to take the various derivatives and plug them into the equation. If you have found the correct solution for y, you will get a true statement. Hope this helps.

 

[Sheneron]

Yes that all helps a lot thank you. One minor question though about the first one.

We are given that y(0) = 6, and we know that y=sin(3t), and when you plug in 0 you get 0. So would the final solution be y= sin(3t) + 6?

 

[EngageEngage]

the general solution for the first one will look like so:

y = c1*sin(3t) + c2*cos(3t) (i only said sin in my earlier post, but you should be able to see that cos would also work). The constants are the coefficients in front of those terms, and you can check to make sure that they work (i didnt check, but I'm pretty sure that's how it should go). So, to find the solution to the IVP, you take the derivative of y to get

y' = 3c1*cos(3t) - 3c2*sin(3t).

now you have to unknowns(c1 and c2) and 2 equations, and 2 intitial problems. from here you just have to plug things in and you will get the values of the constants. Your solution to the ivp will be the solution for y above with the constants being actual values.

 

[Sheneron]

Alright,
Thank you for your help