Linear Partial Differential Equation-Problem in solving?

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SUMMARY

The discussion focuses on solving the linear first-order partial differential equation (PDE) represented by 2p + 3q = x + y + 1. The auxiliary equation derived is dx/2 = dy/3 = dz/(x+y+1). The solution is approached using the method of characteristics and an alternative operator method, leading to the solution z(x,y) = F(y - x*(3/2)) + x^2/8 + xy/2 + x/2, where F(y) is an arbitrary function. The operator method is highlighted as an effective technique for solving this type of PDE.

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Linear Partial Differential Equation---Problem in solving??

Hi guys!

im getting stuck in solving the following Linear Partial Differential Equation:


2p+3q=x+y+1

now when i just form its auxiliary equation it becomes

dx/2 = dy/3 = dz/(x+y+1)

Now the problem that I am just facing is how to solve these three sides. Usually i try to divide and multiply by some constant and then add these three. But in this case, i just couldn't find a way to go!

please help me as soon as possible!

thanks in advance!
 
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You are trying to solve linear first order PDE by the method of characteristics and this way leads certainly to the right answer. However there exist many different methods for solving the problem. I would like to point out the uncommon method - operator method.

If rewrite your equation in the following form

dz/dx =- 3/2*dz/dy + f(x,y)

where f(x,y)=(x+y+1)/2

then its solution in the operator form (See http://arxiv.org/abs/math-ph/0409035 ) follows immediately as

z(x,y) = exp(-x*(3/2)*d/dy)*F(y)+int_0^x[exp(-(x-t)*(3/2)*d/dy)* f(t,y)] dt
where F(y) is an arbitrary function.

Since the operator exp(-x*(3/2)*d/dy) here is the simple shift operator we arrive to

z(x,y) = F(y-x*(3/2))+int_0^x[f(t,y-(x-t)*(3/2))] dt

=F(y-x*(3/2))+x^2/8+xy/2+x/2 .


Yurii
 

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