omarxx84
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Can anyone help me to get the general solution of the linear partial differential equations with variable coefficients of any order?
omarxx84 said: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.
A(x)\frac{\partial^4 Y}{\partial x^4}+ M(x)\frac{\partial^2M}{\partial t^2}= 0omarxx84 said: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.
omarxx84 said:this PDE is correct. but when i use the separation 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 separation 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).
However, I will say once more- if you have two dependent variables to solve for, you will need two equations.