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

Click For Summary
The general solution for the PDE u_{xx} + u = 6y is given by the superposition of the homogeneous solution u(x,y) = Acos(x) + Bsin(x) and the particular solution u(x,y) = 6y, resulting in u(x,y) = Acos(x) + Bsin(x) + 6y. However, this may not be the most general solution, as it could potentially include arbitrary functions of y, such as A(y) and B(y), leading to u(x,y) = A(y)cos(x) + B(y)sin(x) + 6y. The uniqueness of the solution typically depends on boundary conditions or additional assumptions provided in the problem. Without these, it is uncertain whether the derived solution encompasses all possibilities. Therefore, further investigation into the problem's context is necessary to confirm the completeness of the solution.
quantum_smile
Messages
21
Reaction score
1

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?
 
Physics news on Phys.org
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
992
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K