2nd order differential equations I'm so screwed

rockymcrockerso
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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.
 
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You equation is uxx+ 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?
 
HallsofIvy said:
You equation is uxx+ 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).
 
Well, sort of, that form times an arbitrary function of y.

Daniel.
 
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).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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