# Second order differential

1. Dec 1, 2006

### zina_6

I need help solving this equation y''+y'+y= sinx

I know it looks simple but It seems to be getting sticky!! I have been trying to solve it using variation of perimeters,maybe there is a quicker way? If anyone can help please.....

2. Dec 1, 2006

Staff Emeritus

Have you learned how to use Euler's merthod for second order systems with constant coefficients yet?

3. Dec 1, 2006

### zina_6

ty no, not yet Please explain

Last edited: Dec 1, 2006
4. Dec 2, 2006

### ^_^physicist

The form reminds me of a forced oscillator where sin (y) is your forcing function. These aren't too bad. Look at forms along the lines of Ae^yi power as a solution, where i is the imaginary number.

Note e^i*y= cos(y)+i*sin(y), and since the forcing function you are looking for is sin(y) you will want to look at only the imaginary portion of this equation.

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There might be faster way to solve this, but I am going to need to know if you have any intial conditions.

5. Dec 2, 2006

### arildno

Well, the homogenous part of the solution should be easy to find, to find a particular solution, just insert a trial solution $y_{p}=A\cos(x)+B\sin(x)$ and see what happens.

6. Dec 2, 2006

### HallsofIvy

Staff Emeritus
You say you tried variation of parameters so presumably you have already found the general solution to the associated homogenous equation, y"+ Y'+ y= 0. Yes, the solutions to that are a bit complicated:
$$e^{-\frac{x}{2}}\left(C cos(\frac{\sqrt{3}}{2}x)+ D sin(\frac{\sqrt{3}}{2}x)\right)$$
so that while "variation of parameters" is possible, it involves some complicated algebra.

Since the function on the right of the equation, sin(x), is one of the solutions one "expects" for linear equations with constant coefficients, do as arildno suggests- use "undetermined coefficients".

7. Dec 2, 2006

### zina_6

Yes ,That is exactly what i got HallsofIvy, but the integration is very difficult.

8. Dec 2, 2006

### zina_6

I also tried Yp: A cosx + b cos x , after taking the derivatives they seem to cancel out and leave :A sinx + b cosx = sinx A=1, and B=0 it does not seem to work ? I must be missing something or working it wrong?

9. Dec 2, 2006

### zina_6

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L(y)= y''-3y'+2y find L (e^x) and L (Xe^x)

Hence solve the equation y''-3y'+2y=3e^x

This is how I started to work out this problem.

y1=e^x This is what i am assuming to be y1 and y2 ?
y2= Xe^x

then my W = e^2x

Yp = 3x^2e^x/2

Then the general solution Y= c1 e^x + c2 Xe^x + (9X^2e^x)/2

Should I have plugged the value of L (e^x) and L (Xe^x)

Last edited: Dec 3, 2006
10. Dec 2, 2006

### zina_6

OK then lets see what you got !!!!! let me see you break it down for me .

11. Dec 2, 2006

### HallsofIvy

Staff Emeritus
Let y= A sin x+ B cos x
Then y'= A cos x- B sin x
and y"= -A sin x- B cos x

y"+ y'+ y= (A- B- A)sin x+ (B+ A- B)cos x= sin x
or -B sin x+ A cos x = sin x

You must have B= -1, A= 1.

The problem did ask you to solve the differential equation, but only after finding L (e^x) and L (Xe^x). Do that first and perhaps you will see how it helps you solve the equation! The solution you give is not correct. In particular, xex is not a solution to the associated homogeneous equation.

12. Dec 3, 2006

### zina_6

the formula is A cosx + b sinx
yp: A cos x +b sinx
y'p: -A sinx +bcosx
y"p: -A cosx -b sinx

(A-A+b)cosx +(-A+b-b)sinx= sinx
B cosx + -Asinx = sinx
A=-1
b=0

Yp: -1 cosx

y'p : sinx

y"p : cosx

now insert them back in gives you y"+y'+y= sinx
cosx +sinx -cosx =sinx
sinx = sinx