# 3rd Order Non Homogenous diff equation check!

• Nov 2nd 2010, 10:43 AM
FunkyBudduh
3rd Order Non Homogenous diff equation check!
Hi again all,

After solving my last problem i have a new one for you all, ive done it but not sure if it is correct!

$\displaystyle y1(x) = 1, y2(x) = cos 2x, y3(x) = sin 2x.$

Wronskian is 8

Find a (nonhomogeneous) third-order linear di erential equation with general so-
lution using the above

$\displaystyle y(x) = c1y1(x) + c2y2(x) + c3y3(x) + cos x:$

This is what i have,

$\displaystyle (m^2 + 4)(m-1)$

there for the equation is $\displaystyle m^3 - m^2 + 4m -4 =0$

hence Diff Equation is $\displaystyle y - y + 4y -4 = Cos(x)$

thanks

Funky
• Nov 2nd 2010, 11:02 AM
Ackbeet
I think $\displaystyle e^{0}$ is another way to write one of your solutions. That will change your characteristic equation, won't it?
• Nov 2nd 2010, 01:15 PM
FunkyBudduh
no mate, im working backwards from the solutions ive been given to work of the 3rd order differential equations.

thanks

Funkybudduh
• Nov 2nd 2010, 01:24 PM
Ackbeet
Quote:

Im working backwards from the solutions ive been given to work of the 3rd order differential equations.
I realize that. What I'm saying is that if you have a characteristic equation $\displaystyle (m^2 + 4)(m-1)=0,$ then one solution is $\displaystyle m=1$ corresponding to $\displaystyle y=e^{x},$ not $\displaystyle y=1.$ Therefore, I think you need to change your characteristic equation you put forward en route to your DE.
• Nov 3rd 2010, 07:27 AM
FunkyBudduh
im sorry i still dont understand....
• Nov 3rd 2010, 07:56 AM
Ackbeet
Ok. Let me start from a third-order DE, and work the problem in the usual way. I know you're solving the inverse problem of producing a DE, but bear with me. Let's say I have the DE $\displaystyle y'''-8y''+19y'-12y=0.$ The usual procedure with constant coefficient problems like this is to assume a solution of the form $\displaystyle y=e^{mx},$ plug it in, and solve for $\displaystyle m.$ We have $\displaystyle y'=me^{mx}, y''=m^{2}e^{mx},$ and finally $\displaystyle y'''=m^{3}e^{mx}.$ Plugging into the DE yields

$\displaystyle m^{3}e^{mx}-8m^{2}e^{mx}+19me^{mx}-12e^{mx}=0.$ Factoring yields

$\displaystyle e^{mx}(m^{3}-8m^{2}+19m-12)=0.$ Now $\displaystyle e^{mx}\not=0,$ hence we may divide out to obtain the characteristic equation

$\displaystyle m^{3}-8m^{2}+19m-12=0.$ Now you can show that the roots of this equation are $\displaystyle m=4,3,1.$ That is, it factors as $\displaystyle (m-4)(m-3)(m-1)=0.$ Hence, the solution to this DE is as follows:

$\displaystyle y=c_{1}e^{4x}+c_{2}e^{3x}+c_{1}e^{x}.$

Do you see (and this is really where I'm going with this), that the root $\displaystyle m=1$ corresponds, because of our initial guess of $\displaystyle y=e^{mx}$, to the solution $\displaystyle y=e^{x}?$ It does NOT correspond to the solution $\displaystyle y=1.$

If you don't believe me, think of it this way: according to the OP, you are claiming that $\displaystyle y=1$ solves the homogeneous ODE

$\displaystyle y''' - y'' + 4y' -4y =0.$

I altered your DE with some obvious corrections. Plug $\displaystyle y=1$ into this DE, and you will see that it does not satisfy it.

• Nov 3rd 2010, 09:14 AM
FunkyBudduh
Ah now i get you, sorry about that. Well back to the drawing board, ill let you know if i have anything. Will probably start with a trig function and do it the way youve just showed me.

thanks

Funky
• Nov 3rd 2010, 09:57 AM
Ackbeet
I gave you a decent hint in Post # 2. What could post # 2 tell you?
• Nov 4th 2010, 04:35 AM
FunkyBudduh
Then my new characterisctic equation is

$\displaystyle m(m^2 + 4)=0$

hence my 3rd order diff equation is

$\displaystyle d^3y/dx^3 + 4dy/dx = g(x)$

but how would i use this to find g(x) function?

if the general solution of the 4rd order linear de (non homog) is
$\displaystyle y(x)=c1y1(x) + c2y2(x) + c3y3(x) +Cos(x)$

where y1, y2 and y3 are 1, cos(2x), sin(2x)
• Nov 4th 2010, 04:48 AM
Ackbeet
I agree with your DE now.

Well, you know what the LHS operator $\displaystyle \frac{d^{3}}{dx^{3}}+4\frac{d}{dx}$ does to the homogeneous part of your starting solution $\displaystyle c_{1}y_{1}+c_{2}y_{2}+c_{3}y_{3}.$ Find out what it does to the $\displaystyle \cos(x)$ part. Then just use that to determine the RHS of the DE. Be careful with the signs!
• Nov 4th 2010, 06:18 AM
FunkyBudduh
ok, my answer for g(x)= -3sin(x)

ive taken cos(x) and found dy/dx and d^3/dx^3 then substituted it in dy^3/dx^3 + 4dy/dx = g(x). which yields g(x) = -3Sin(x)

thanks again ackbeet
• Nov 4th 2010, 06:31 AM
Ackbeet
Yep, that looks right. You're welcome! Have a good one.