# ODEs with variable coefficients

1. May 14, 2010

### omarxx84

Can anyone help me to get the general solution of the linear partial differential equations with variable coefficients of any order?

2. May 14, 2010

### phyzguy

No one knows how to do that. PDEs are quite complicated.

3. May 14, 2010

### omarxx84

can anyone help me to solve this ODE:
A(x)Y''''(x,t)+M(x)W''(x,t)=0
where: A(x) and M(x) are variable coefficients
Y''''(x,t) 4th derivative with respect to x.
w''(x,t) 2nd derivative with respect to t.

4. May 14, 2010

### jasonRF

That is a PDE, not an ODE. What is A, what is M, what is Y, what is W? What are boundary/initial values? Unless you get specific, espcially regarding A and M, no one can help you.

jason

5. May 15, 2010

### omarxx84

this PDE is correct. but when i use the seperation of variables principle to solve this equation, the resulting two equations are ODEs, which may as follows:
A(x)Y''''(x,t)+M(x)Y''(x,t)=0 ........(1) (main equation)
PUT : Y=F(x)W(t) AND SUBSITITUTING IN THE ABOVE EQUATION AND SEPERATING THE VARIABLES, WE GET THE TWO ODEs AS FOLLOWS:
W''(t)-aW(t)=0 ..........(2)
where: a is arbitrary constant. this may be solved easily.and the second ODE is:
{A(x)/M(x)}F''''(x)-aF(x)=0 .........(3)
this equation is linear ODE with variable coefficient.
in which: A(x) and M(x) are variable coefficients along x-axis.
a is arbitrary constant, from seperation of variable process.
by solving equation(3) and compiling with the solution of equation(2), we will get the solution of equation(1). so, the problem is how we can solve the equation(3). boundary or initial conditions, i don't we need now to solve equation(3).

6. May 15, 2010

### HallsofIvy

Staff Emeritus
$$A(x)\frac{\partial^4 Y}{\partial x^4}+ M(x)\frac{\partial^2M}{\partial t^2}= 0$$
Not only do you have two independent variables, x and t, you have two dependent variables, Y and M. One equation is not sufficient to solve for two unknown functions.

Last edited: Jun 2, 2010
7. May 16, 2010

### jackmell

What you seemed to want was:

$A(x)\frac{\partial^4 Y}{\partial x^4}+M(x)\frac{\partial^2 Y}{\partial t^2}=0$

and by letting $Y(x,t)=F(x)W(t)$ and separation constant equal to a, you get one ODE in t and another ODE in x. You then want to know how to solve the ODE in x:

$\frac{d^4 F}{dx^4}=b(x)F$

where $b(x)=\frac{aM(x)}{A(x)}$

Here's what I'd do with this and every other problem like it: First just try it in Mathematica or Wolfram Alpha:

DSolve[y''''[x] == b[x] y[x], y, x]

who knows, might get an answer and then that answer could conceivably help you derive an analytical method to obtain it. However in this case, Mathematica can't come up with one. Doesn't mean some nice method can't find it but rather, just Mathematica. However if A(x) is a constant and M(x) is a polynomial, then looks like you could solve it via simple power series. Additionally if b(x) is analytic with power series $b(x)=\sum c_n x^n$ you could still conceivably solve it via power series but would need to form the Cauchy Product of the resulting series product. For example, the equation $y''''-e^x y=0$ is not too hard to solve this way for say x in (0,1). See "Intermediate Differential Equations" by Rainville for a (simple) example.

Last edited: May 16, 2010
8. May 21, 2010

### omarxx84

i want to ask. what are the "differential forms"? and what are the distinct applications?

9. Jun 2, 2010

### omarxx84

Hi colleages. can you help me to solve the cubic equation below:
2N(Ep-En)hp^3(x)-3[M(x,t)(Ep-En)-2NEnh]hp^2(x)-6Enh[M(x,t)+Nh]hp(x)+Enh^2[3M(x,t)+2Nh]=0 notice that all variables in the equation are dependent on x only, except M is dependent on x and t.
En, Ep, N and h are constants.
i know the solution when the equation dependent on one variable, but the problem is that M dependent on two variables x and t.

10. Jun 12, 2010

### omarxx84

CAN ANYONE GIVE ME A GUIDANCE TO SOLVE THE EQUATION BELOW:

EI(x,t)U''''(x,t)+M(x,t)V''(x,t)=0
where
U''''=4th partial derivative with respect to x only.
V''=2nd partial derivative with respect to t only.

11. Jun 18, 2010

### omarxx84

please, my colleages, can tell me is the equation below is linear or nonlinear PDE?
EI(x,t)U''''(x,t)+M(x,t)V''(x,t)=0
U''''=4th partial derivative with respect to x only.
V''=2nd partial derivative with respect to t only.

12. Jun 18, 2010

### HallsofIvy

Staff Emeritus
Since you have no powers or more complicated functions of the dependent variables U and V, that is a linear equations. However, I will say once more- if you have two dependent variables to solve for, you will need two equations.

13. Jun 18, 2010

### KrayzBlu

$$\Delta$$u(x,y) = u(x,y)*f(x,y)