# Thread: 3rd Order Non Homogenous diff equation check!

1. ## 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!

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

Wronskian is 8

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

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

This is what i have,

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

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

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

thanks

Funky

2. I think $e^{0}$ is another way to write one of your solutions. That will change your characteristic equation, won't it?

3. no mate, im working backwards from the solutions ive been given to work of the 3rd order differential equations.

thanks

Funkybudduh

4. 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 $(m^2 + 4)(m-1)=0,$ then one solution is $m=1$ corresponding to $y=e^{x},$ not $y=1.$ Therefore, I think you need to change your characteristic equation you put forward en route to your DE.

5. im sorry i still dont understand....

6. 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 $y'''-8y''+19y'-12y=0.$ The usual procedure with constant coefficient problems like this is to assume a solution of the form $y=e^{mx},$ plug it in, and solve for $m.$ We have $y'=me^{mx}, y''=m^{2}e^{mx},$ and finally $y'''=m^{3}e^{mx}.$ Plugging into the DE yields

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

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

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

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

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

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

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

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

7. 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

8. I gave you a decent hint in Post # 2. What could post # 2 tell you?

9. Then my new characterisctic equation is

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

hence my 3rd order diff equation is

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

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

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

where y1, y2 and y3 are 1, cos(2x), sin(2x)

10. I agree with your DE now.

Well, you know what the LHS operator $\frac{d^{3}}{dx^{3}}+4\frac{d}{dx}$ does to the homogeneous part of your starting solution $c_{1}y_{1}+c_{2}y_{2}+c_{3}y_{3}.$ Find out what it does to the $\cos(x)$ part. Then just use that to determine the RHS of the DE. Be careful with the signs!

11. 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

12. Yep, that looks right. You're welcome! Have a good one.