# Ordinary Differential Equations (ODE) Problem

1. Aug 31, 2015

### Mark Brewer

1. The problem statement, all variables and given/known data
dy/dx + 2sin2pix = 0 -------Answer: y = 1/pi cos2pix + c

2. Relevant equations

3. The attempt at a solution
I made several attempts but no success to the correct answer. The first step I made was subtracting 2sin2pix to both sides. I then used integration by parts, and this is where I've stopped because I was confused on how might my answer be anything like the true answer. This is where I stopped: x/pi cos2pi - integral 1/pi cos2pi dx.

Any help would be much appreciated.

2. Aug 31, 2015

### Staff: Mentor

Have you tried differentiating the answer to see what you get?

3. Aug 31, 2015

### Kilgour22

Instead of using integration by parts, is there another method we can use, such as letting $u = 2\pi{x}$, then getting your integral with respect to $du$ instead of $dx$?

EDIT: Bonus question: Can you spot the flaw with using integration by parts for this integral, and why it will not yield anything useful?

Last edited: Aug 31, 2015
4. Aug 31, 2015

### Staff: Mentor

Mark: If you had to integrate sinθ dθ, would you integrate by parts?

Chet

5. Aug 31, 2015

### Mark Brewer

Thank you for the replies.

Chestermiller, I would not use integration by parts. I would simply integrate.

Kilgour22, thank you for the hint. I didn't think about using 2πx in u-substitution. I figured that was attached to the sine function.

I really appreciate all your help. I hope this wasn't a hassle. I do plan on using this forum throughout my undergraduate.

Cheers,

Mark

6. Aug 31, 2015

### zomorodian

Dear Mark. Take 2sin2pix to the other side so it will become like this:
dy/dx=-2sin2pix. then multiply both sides to dx, it will become like this:
dy=-2sin2pix(dx). Now just a simple integration will do the job for you.
u just need to use u=2pix, then create du and the rest is easy for u I'm sure.
cheers

7. Aug 31, 2015

### Kilgour22

No worries! It just takes practice, and eventually you'll become so familiar with integration that the answer to such problems will be solvable in your head rather easily. :)

8. Aug 31, 2015

### Mark Brewer

I've got another ODE problem.

dy/dx = 4e-xcosx

I've divided dx to both sides, and now have dy = 4e-xcosx dx
I've then started to use intergration by parts to the right side with u = 4e-x and dv = cosx dx
Leaving y = 4e-xsinx - ∫ -4e-xsinx dx
Once again I used intergration by parts with u = -4e-x and dv = sinx dx
Leaving y = 4e-xsinx - (4e-xcosx - ∫ -4e-xcosx dx)
I know that I don't need to integrate again, so my final answer is 4e-x(sinx - cosx) + c

The correct answer is 2e-x(sinx - cosx) + c

Where am I missing division by 2?

Cheers,

Mark

9. Sep 1, 2015

### MidgetDwarf

Typically for these simple cases, if you can isolate the different dependent variables on on side do it. Then if you can also keep the indipendent variables with there similar dependent variables do so.

ie dy/dx-e^x=0
dy/dx= -e^x

then multiply the dx on both sides of the equation.

dy=-e^x(dx)

integrate both sides with respectively for dy and dx

y=-e^x +c

This work for simpler cases.

10. Sep 1, 2015

### HallsofIvy

Staff Emeritus
Mark Brewer, you understand that all of these "Ordinary Differential Equation" problems you have posted are just integration problems in disguise, don't you?