Simple PDE: Finding the General Solution for u_{xx} + u = 6y

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SUMMARY

The general solution for the partial differential equation (PDE) \( u_{xx} + u = 6y \) is expressed as \( u(x,y) = A \cos(x) + B \sin(x) + 6y \), where \( A \) and \( B \) are arbitrary constants. The homogeneous solution is \( u_h(x,y) = A \cos(x) + B \sin(x) \) and the particular solution is \( u_p(x,y) = 6y \). To ensure this is the most general solution, one must consider the possibility of \( A(y) \) and \( B(y) \) being arbitrary functions of \( y \), which would yield a broader solution set.

PREREQUISITES
  • Understanding of partial differential equations (PDEs)
  • Knowledge of homogeneous and particular solutions
  • Familiarity with linear independence in solutions
  • Basic concepts of boundary conditions in differential equations
NEXT STEPS
  • Explore the method of superposition in solving PDEs
  • Learn about boundary value problems and their role in determining unique solutions
  • Investigate the implications of arbitrary functions in solutions, specifically \( A(y) \) and \( B(y) \)
  • Study examples of PDEs with varying boundary conditions to understand solution uniqueness
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Students and professionals in mathematics, particularly those studying differential equations, as well as educators looking for clear examples of solving PDEs and understanding solution generality.

quantum_smile
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Homework Statement


Find the general solution of
u_{xx} + u = 6y,
in terms of arbitrary functions.

Homework Equations



The PDE has the homogeneous solution, u(x,y)=Acos(x)+Bsin(x).
u_{xx} + u = 6y has the particular solution, u(x,y)=6y

The Attempt at a Solution



Taking a superposition of the homogeneous and particular solutions, we can write that
u(x,y)=Acos(x)+Bsin(x)+6y. <--my solution

My question is, how do I know whether or not this is *the* most general solution? How do I know that I haven't missed something?
 
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quantum_smile said:

Homework Statement


Find the general solution of
u_{xx} + u = 6y,
in terms of arbitrary functions.


Homework Equations



The PDE has the homogeneous solution, u(x,y)=Acos(x)+Bsin(x).
u_{xx} + u = 6y has the particular solution, u(x,y)=6y

The Attempt at a Solution



Taking a superposition of the homogeneous and particular solutions, we can write that
u(x,y)=Acos(x)+Bsin(x)+6y. <--my solution

My question is, how do I know whether or not this is *the* most general solution? How do I know that I haven't missed something?

You know that ##y_h## and ##y_p## are both linearly independent solutions to the equation. Therefore the sum of those two linearly independent solutions is also a linearly independent solution. Hence the most general solution (although not necessarily unique) is ##y = y_h + y_p##.

Consider a regular ODE for a moment of the form: ##a(x)y'' + b(x)y' + c(x)y = f(x)##.

Differentiating ##y## twice and plugging it into the above equation you will find the answer to be ##0 + f(x) = f(x)##.
 
quantum_smile said:

Homework Statement


Find the general solution of
u_{xx} + u = 6y,
in terms of arbitrary functions.

Homework Equations



The PDE has the homogeneous solution, u(x,y)=Acos(x)+Bsin(x).
u_{xx} + u = 6y has the particular solution, u(x,y)=6y

The Attempt at a Solution



Taking a superposition of the homogeneous and particular solutions, we can write that
u(x,y)=Acos(x)+Bsin(x)+6y. <--my solution

My question is, how do I know whether or not this is *the* most general solution? How do I know that I haven't missed something?

You don't know that it is. And obviously it isn't because you could have$$
u(x,y) = A(y)\cos x + B(y)\sin x + 6y$$where ##A(y)## and ##B(y)## are arbitrary functions of ##y##. Is this the most general solution? It might be, but I don't know. Usually problems like this come with some boundary conditions and assumptions which guarantee a unique solution. When you have that situation, then if you come up with something that works, no matter how you found it, you know you are done because you have the only solution there is.
 

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