2nd order differential equations I'm so screwed

[rockymcrockerso]

Homework Statement

Uxx +u(x,y)=0

Homework Equations

?

The Attempt at a Solution

Step 1) u(x,y)=A(x)B(y)
Step 2) uxx=d^u/dx^2
Step 3) d^u/dx^2 + A(x)B(y) = 0
Step 4) d^u/dx^2 = -[A(x)B(y)] (?)

I have no idea what I'm doing, so small words would be useful.

 

[HallsofIvy]

You equation is u_xx+ u(x,y)= 0?
$$\frac{\partial^2u}{\partial x^2}+ u= 0$$

Since there is no differentiation with respect to y, just treat y as a constant. Can you solve u'+u= 0?

 

[rockymcrockerso]

You equation is u_xx+ u(x,y)= 0?
$$\frac{\partial^2u}{\partial x^2}+ u= 0$$

Since there is no differentiation with respect to y, just treat y as a constant. Can you solve u'+u= 0?

Yeah I figured it out...it's Acosx+Bsinx...right? (I warned that I don't know what the heck I'm doing).

 

[dextercioby]

Well, sort of, that form times an arbitrary function of y.

Daniel.

 

[HallsofIvy]

No, not "that form times an arbitrary function of y".

Since y is being treated as a constant, the constants A and B may be functions of y: u(x,y)= f(y)cos(x)+ g(y)sin(x). That is not quite the same thing. For example, f(y)cos(x) satisfies the differential equation but is not a function of y times cos(x)+ sin(x).