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Here is the PDE : $U^2_x + U^2_y + 1 = \frac{1}{U^2}$

I tried to solve it using separation of variables method. Assume $U=XY$; $U_x = \dot X Y$ and $U_y = X \dot Y$

so the PDE become : $(\dot X Y)^2 + (X \dot Y)^2 + 1 = \frac{1}{(XY)^2}$

$(\dot X Y)^2 + (X \dot Y)^2 = \frac{1}{(XY)^2} - 1$

My goal is to make the RHS become $0$ however i am stuck.

Ocean
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  • what if you solve the homogenous and then assume the particular is proportional to it? – hellofriends Mar 26 '24 at 05:46
  • do you mean i should try to solve the $(\dot X Y)^2 + (X \dot Y)^2 = 0$ first? – Ocean Mar 26 '24 at 05:49
  • Separation of variables does not help. If you divide your final step by $X^2Y^2$, you get$$\frac{\dot{X}}{X}+\frac{\dot{Y}}{Y}=\frac{1}{(XY)^4}-\frac{1}{(XY)^2}.$$Taking the partial derivative of each side with respect to $X$, say, will give you a pure function of $X$ on the left, but not on the right, so no solutions of the form $XY$ exist. – Theo Bendit Mar 26 '24 at 08:12
  • Homogeneous solution: $U! \left(x,y\right)=f_{1}! \left(y-\mathrm{I} x\right)+f_{2}! \left(y+\mathrm{I} x\right)-\frac{x^{2}}{2}$ – gpmath Mar 26 '24 at 09:08
  • What kind of solution are you looking for? A complete integral or a solution satisfying some boundary condition? – Gonçalo Mar 26 '24 at 15:23

1 Answers1

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Hint.

Using a Charpit procedure we have:

$$ f(x,y,z,p,q) = p^2+q^2+1-\frac{1}{z^2} $$

and from

$$ \frac{dp}{f_x+p f_z}=\frac{dq}{f_y+q f_z}=\frac{dz}{-p f_p-q f_q}=\frac{dx}{-f_p}=\frac{dy}{-f_q} $$

we have

$$ \frac{z^3dp}{p}=\frac{z^3dq}{q}=-\frac{dz}{p^2+q^2}=-\frac{dx}{p}=-\frac{dy}{q} $$

From

$$ \frac{z^3dp}{p}=\frac{z^3dq}{q}\Rightarrow p = c_1 q $$

From

$$ \frac{z^3dp}{p}=-\frac{dz}{p^2+q^2}\Rightarrow \frac{dz}{z^3}=-(c_1^2+1)q dq $$

etc.

Cesareo
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