# Initial Value Problem (complex example)

1. Dec 3, 2015

### King_Silver

I know the method and can solve other initial value problems. This is the question given:
dy/dx + y(-2) Sin(3x) = 0 for t > 0, with y(0) = 2.

I've brought the dy/dx and let it equal to the rest of the expression so it is now:
dy/dx = -y-2 Sin(3x) , with y(0) = 2 (i.e. when x = 0, y = 2 )

The -y-2 is obstructing me from integrating both sides and solving for constant C using the information given. What would I have to do to eliminate the -y-2 entirely from the question so I can continue standard procedures?

2. Dec 3, 2015

### Krylov

Maybe I'm overlooking something, but can't you just multiply the original equation by $y^2$?

3. Dec 3, 2015

### King_Silver

But if you multiply the original by y2 then you will need to do the same to the left hand side as you have done to the right hand side, meaning the dy/dx would also be affected.

4. Dec 3, 2015

### SteamKing

Staff Emeritus
It's not clear which is the dependent variable here. The equation suggests x, but there's this 't > 0' restriction. I'm confused.
It looks like you want to use separation of variables here. Do you know how to get all y and dy to one side of the equation, and all x and dx to the other?
If you separate the variables, everything should be smooth sailing after that.

5. Dec 3, 2015

### King_Silver

What do you mean by that? all y to the left and all x terms to the right?

so like ∫ y-2 dy = ∫Sin(3x) dx ??? :)

6. Dec 3, 2015

### Krylov

Indeed. Just allow yourself to do the usual abuse of differentials.

7. Dec 3, 2015

### SteamKing

Staff Emeritus
I think you have too many integral signs on one side of the equation, and not enough on t'other.

8. Dec 3, 2015

### King_Silver

Sorry that was just an accident edited the post :) thanks! got it sorted now cheers! :D

9. Dec 3, 2015

### Samy_A

Shouldn't that be ∫ y+2 dy?

10. Dec 3, 2015

### King_Silver

And a -Sin(3x) I think, sorry my bad!!!
I'm sure it should be a -Sin(3x) as well.

11. Dec 3, 2015

### King_Silver

integrated it.
y3/3 = 1/3 Cos(3x) + C, with y(0) = 2

so 23/3 = 1/3 Cos(0) + C
2 2/3 = 1/3 + C

C = 2 1/3

That is what I got :)

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