MHB Partial DE-separation of variables

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Hi I'm having a bit of trouble with this question:

Use separation of variables to find all the possible separable solutions to the partial DE equation for u(x,y) given by
yux - 3x2 uy = 0
.I try u= X(x) Y(y)

ux = X'(x) Y(y)

uy = X(x) Y'
(y)

which gives y(X' Y)-3x2(X Y')

then I divide by (XY) and rearrange it into y[X'/X] = 3x2 [Y'/Y]

which is equal to the separationconstant K (I think)

y[X'/X] = 3x2 [Y'/Y] = Kand from here I get a bit lost. I try a positive constant, so I get two equations.

yX'/X = + K rearranged into {X'= XK/y}

3x2Y/Y = + K rearranged into {Y'= YK/3x2 }

and then I try a negative constant with more or less the same equations resulting, but I'm not sure what to do with them?

Is this how you find all the possible solutions of the pde (trying both a positive and a negative constant)?

I'm told I can check the solution by substitution but I'm not sure how. Any help would be really appreciated.

Thanks:confused::confused:.
 
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emily600 said:
Hi I'm having a bit of trouble with this question:

Use separation of variables to find all the possible separable solutions to the partial DE equation for u(x,y) given by
yux - 3x2 uy = 0
.I try u= X(x) Y(y)

ux = X'(x) Y(y)

uy = X(x) Y'
(y)

which gives y(X' Y)-3x2(X Y')

then I divide by (XY) and rearrange it into y[X'/X] = 3x2 [Y'/Y]

which is equal to the separationconstant K (I think)

y[X'/X] = 3x2 [Y'/Y] = K

You would do better to rearrange it as $\mathbf{\dfrac{X'}{x^2X}} = \mathbf{\dfrac{3Y'}{yY}}$. The left side is then purely a function of $\mathbf{x}$ and the right side is then purely a function of $\mathbf{y}.$ You are then justified in putting both sides equal to a separation constant, and you have two separate ODEs for $\mathbf{x}$ and $\mathbf{y}$.
 
Thanks that simplified things a little. For the positive constant I solved the 2 ODE's as:

View attachment 1591

And then put them together. If I do the same for a negative constant would that cover the possible solutions (assuming I can rearrange and solve them properly)?

Thanks:D
 

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I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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