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Given PDE is $$x\frac{\partial u}{\partial x}+(u-x-y)\frac{\partial u}{\partial y}=x+2y$$ Find integral surface which also passes through $u=1+y$ on $x=1$.

I used Lagrange's Auxiliary equations & found $$\frac{x+y+u}x=A, \frac{x+y+u}{y+u}=B$$ I got $A$ & $B$ relation as $B = \dfrac A{A-1}$ by using curve given.

While finding last integral equation (by substituting $A$ & $B$ expressions) I am unable to get an integral surface. Instead I am getting $1 = 1$.

Please tell me, where am I making mistake?

user170231
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Dvaavimshathihi
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Apr 05 '23 at 13:57
  • @ Dvaavimshathihi. You did a good work on the way to solve the problem. Your two characteristic equations are both correct but unfortunately they are not undependent one to the other. See my answer. – JJacquelin Apr 05 '23 at 18:25

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The two characteristic equations that you found are correct. This is a good start. $$\frac{x+y+u}{x} = A$$ $$\frac{x+y+u}{y+u} = B$$ The hitch is that they are not undependent because $$\frac{y+u}{x}=\frac{A}{B}=A-1$$ Of course one can keep the simplest characteristic equation $$\frac{y+u}{x}=c_1$$ But one have to find another undependent characteristic equation from the Charpit-Lagrange ODEs. $$\frac{dx}{x}=\frac{dy}{u-x-y}=\frac{du}{x+2y}$$ $u=c_1x-y\quad\implies\quad \frac{dx}{x}=\frac{dy}{(c_1x-y)-x-y}\quad\implies\quad \frac{dy}{dx}=c_1-1-\frac{2y}{x}$

Solving this first order linear ODE leads to $y=\frac{c_1-1}{3}x+\frac{c_2}{x^2}$

A second characteristic equation is $y=\frac{\frac{y+u}{x}-1}{3}x+\frac{c_2}{x^2}$ and after simplification with $c_3=-2c_2$ : $$u=2y+x+\frac{c_3}{x^2}$$

The general solution of the PDE on implicit form is : $$u(x,y)=2y+x+\frac{1}{x^2}F\left( \frac{y+u}{x}\right)$$ $F$ is an arbitrary function to be determined according to the condition $u(1,y)=y+1$.

I suppose that you can do it. My result is :

$$u(x,y)=\frac{x-y+2x^3(2y+x)}{2x^3+1}$$ This is the particular solution which satisfies both the PDE and the condition. To be checked.

JJacquelin
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