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Is there a method to solve this in a simple way(For a multiple choice question).

Question is, If $u$ is a function defined on $R^2$ and satisfies the equation $\partial^2U/\partial x^2$-$\partial^2U/\partial y^2$ then the general solution of the equation is:
a)u(x,y)= $ 2f(x+2y)+g(x-2y) $
b)u(x,y)= $(x+y)f(x-y)+g(x-y) $
c)u(x,y)= $(x+2y)f(x+y)+g(x-y)$
I know how to solve these normally, but it takes time for a mcq.Is there a method I can follow to gain the answer quickly

clarkson
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2 Answers2

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Hint: You can use separation of variables techniques which is assume the solution has the form

$$ U(x,y)= F(x)G(y). $$

See a related technique.

Community
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Mhenni Benghorbal
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Are you sure you did not misspelled the PDE or the answers? The solution to the PDE is:

$$U(x,y) = f(x+y) + g(x-y),$$ where $f$ and $g$ are arbitrary functions of its arguments and $\xi = x+y$ and $\eta = x-y$ are the characteristics of the equation.

This solution can be obtained by several ways. For example, by using the D-Operator. Indeed, writing the PDE as follows:

$$(D^2_x - D^2_y) U = 0,$$

where $D^2_{x_i} = \frac{\partial^2}{\partial x_i^2}, \, x_i = \{x,y\}$, you can make the following decomposition:

$$(D_x - D_y)(D_x + D_y)[U] = 0, \quad (D_x-D_y)[\omega] = 0, $$

leading to a 1st order PDE for $\omega$, whose solution can be obtained in terms of the characteristics method:

$$ dx = dy = \frac{d\omega}{0} $$

which yields: $\omega(x,y) = F(x+y)$.

Then, by definition:

$$ (D_x + D_y) U = \omega = F(x,y), $$

which readily yields to the solution shown above.

Cheers!

Dmoreno
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