# 4th order equation

1. Apr 11, 2007

### ranger1716

I have a question regarding a 4th order differential equation from an exam i just took.

we were asked to solve y^(4)-1=5 given y'(0)=y''(0)=y^(3)(0)=0

I started by factoring down to (r-1)(r+1)(r^2+1)=5.

I then found my general solution to be y=C_1e^6x+C_2e^4x+C_3e^2x+C_4e^-2x

Obviously I would then be left with four equations with four unknowns to solve for my constants. Would I need to use a solver and/or hand solve the equations in order to find the constants? I didn't have time to do that so I just put the equations into a matrix and said that the constants were all equal to zero.

Just thought I would ask what the right approach would be (a little to anxious to wait another week)

2. Apr 11, 2007

### Data

Can you clarify what your DE is:

Is it $y^{(4)} - y = 5$, or $y^{(4)} - 1 =5$?

In the second case, rearrange it to $y^{(4)} = 6.$ The answer is just some 4th order polynomial (the coeffs are easy so I'll let you work them out. Think about it for a minute ).

In the first case the answer is even simpler; y(x) = -5.

Edit: Ignore this and see below.

Last edited: Apr 11, 2007
3. Apr 11, 2007

### AiRAVATA

If what you wrote is correct, then the solution is $y(x)=\frac{x^4}{4}+c$, where $c$ is a constant.

If what you meant is $y^{(4)}-y=5$, then you are missing one condition on $y(0)$, and your solution is incorrect.

What went wrong you may ask. Well, the solution of the second equation is the sum of the homogeneous equation $y_h^{(4)}-y_h=0$, plus the solution of the particular equation $y_p^{(4)}-y_p=5$. The particular solution is clearly $y_p=-5$, so you only have to solve the homogeneous equation, which indeed has the form

$$y_h(x)=c_1e^{r_1x}+c_2e^{r_2x}+c_3e^{r_3x}+c_4e^{r_4x},$$

where $r_i$ are the roots of the characteristic equation $r^4-1=0$. Therefore, the solution is

$$y_h(x)=c_1e^{x}+c_2e^{-x}+c_3e^{ix}+c_4e^{-ix},$$

and evaluating on the initial conditions,

$$y(x)=\frac{y_0+5}{2}\left[\cosh x+\cos x\right]-5,$$

where $y(0)=y_0$.

Last edited: Apr 11, 2007
4. Apr 11, 2007

### Data

Whoops, you're right. I forgot about the missing fourth init. cond.!

Last edited: Apr 11, 2007
5. Apr 11, 2007

### AiRAVATA

Lets hope the missing condition is $y(0)=-5$ ;)

6. Apr 11, 2007

### Data

Indeed! I was about to post the problem with regard to the 2 in front of the cos term (I needed to check that I wasn't making another dumb mistake, first!), but I see that you've already gotten to it.

7. Apr 11, 2007

### AiRAVATA

Lol, I always have to edit my posts like 7 times :P

8. Apr 11, 2007

### Data

Yes, it seems like a common phenomenon with mine as well .