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I'm trying to solve the following equation:

$$\frac{1}{e^x\sqrt{y} + 1 } U_x + yU_y = y^2u \quad \text{u, y>0}$$

my try:

Solving with Lagrange's method - I wrote the characteristic lines equation: $$ dx(e^x\sqrt{y} + 1) = \frac{dy}{y} = \frac{du}{uy^2} \tag 1 $$

I went with the easier looking one to begin with: $$ \frac{dy}{y} = \frac{du}{uy^2} \quad \implies \quad ydy = \frac{du}{u} \quad \implies \quad \frac{1}{2}y^2 +C = ln|u|$$ which gives the first surface: $$\phi_1 = u\cdot e^{-\frac{1}{2}y^2} $$

Now here's where I got stuck, solving the second equation I went with $dx$ and $dy$ equations, (I also tried different approaches but none gave better results:

$$ \tag 2 dx(e^x\sqrt{y} + 1) = \frac{dy}{y}$$ And I don't know how to reach a solution from here, my latest try was: Since y is not a constant => Isolating y from $\phi_1$ and substituting in eq (2): $\sqrt{y} = (2\phi_1 + 2ln|u|)^\frac{1}{4}$

$$ \int e^x(2\phi_1 + 2ln|u|)^\frac{1}{4} + 1 dx = \int \frac{dy}{y} = ln|y| \tag 3$$

but the same problem remains, u is not a constant so I cannot integrate $\int dx$. Help would be greatly appreciated!

the solution to this problem: $$ u(x,y) = e^{\frac{y^2}{2} - f(\frac{1}{\sqrt{y}}e^\frac{x}{2} + \frac{1}{3}e^{\frac{3x}{2}})} $$

Edward Josef
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  • Are you sure the problem is stated correctly? I don't see a nice solution to the $dx/dt=(y_0^{1/2} e^x e^{t/2}+1)^{-1}$ equation that is obtained to develop the characteristics. – Ian Oct 20 '21 at 17:24
  • @lan Yes, the only thing I failed to mention (edited now) is that u>0 y>0, sorry :) – Edward Josef Oct 20 '21 at 17:27
  • Yeah I'm pretty sure you need to be able to solve $dx/dt=(y_0^{1/2} e^x e^{t/2}+1)^{-1},x(0)=x_0$ in order to solve this, and the solution I get out of that from Wolfram Alpha is not pretty. – Ian Oct 20 '21 at 17:38

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You can parametrize the characteristics using just $x$ in this situation. The reason is that if you parametrize by $x$ then you get $\frac{dy}{dx}=e^x y^{3/2} + y$ or equivalently $\frac{dy}{dx}+(-1)y=e^x y^{3/2}$. This is a Bernoulli equation that you can solve by starting with $v=y^{-1/2}$ which yields $\frac{dv}{dx}=-\frac{1}{2} (e^x + v)$ which is first order linear. Now I think you can probably turn the crank to develop the characteristics and then determine $u$ along each characteristic.

Ian
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