The characteristic curves of PDE
$(2x+u)u_x + (2y+u)u_y = u$
passing through $(1,1)$ for any arbitrary initial values prescribed on a non characteristic curve are given by:
- $x=y$
- $x^2 + y^2 = 2$
- $x + y = 2$
- $x^2 + y^2 - xy = 1$
It's a single select question, that is exactly one the above options is true which gives the characteristic curves. Pardon if it seems trivial but I am not aware of the steps that shall be involved in solving this question. Kindly help out, thanks in advance.
In comment you wrote : " I suppose to learn the concept from this question if it gets answered here". I think this is not a good idea. Taking the problem backward doesn't simplify the understanding. If you want learn from examples, better look at examples of direct solving of PDEs. They are a lot of examples fully solved on the forum.
Nevertheless I post a method of solving below. But I doubt that this will help you much to "understand the concept" as you say.
$$(2x+u)u_x+(2y+u)u_y=u $$ The Charpit-Lagrange characteristic ODEs are : $$\frac{dx}{2x+u}=\frac{dy}{2y+u}=\frac{du}{u}$$ A first characteristic equation comes from solving $\frac{dx}{2x+u}=\frac{du}{u}$ leading to : $$\frac{x+u}{u^2}=c_1$$ A second characteristic equation comes from solving $\frac{dy}{2y+u}=\frac{du}{u}$ leading to : $$\frac{y+u}{u^2}=c_2$$ The general solution of the PDE expressed on the form of implicit equation $\Phi(c_1,c_2)=0$ is : $$\Phi\left(\frac{x+u}{u^2}\:,\:\frac{y+u}{u^2}\right)=0$$ $\Phi$ is an arbitrary function of two variables.
This is a method to solve a problem when a boundary condition is specified, allowing to determine the function $\Phi$. This is not the case of the present problem. Thus the above general solution is of no use.
To go further in the present case we note that they are a lot of characteristic equations dependant one from others which can be written in combining the two above ones. For example we look for some characteristic equations involving only $x$ and $y$ without $u$.
$\frac{c2}{c1}=\frac{y+u}{x+u}\quad\implies\quad u=\frac{c_2x-c_1y}{c_1-c_2}$
$c_1=\frac{x+u}{u^2}=\frac{x+\frac{c_2x-c_1y}{c_1-c_2}}{(\frac{c_2x-c_1y}{c_1-c_2})^2}\quad$ and after simplification : $$(c_2x-c_1y)^2+(x-y)(c_2-c_1)=0$$ Now we look if among the four characteristic equations proposed in the multiple choice, one of them is consistent with $(c_2x-c_1y)^2+(x-y)(c_2-c_1)=0$ for some specific values of $c_1$ and $c_2$.
We observe that the cases 2,3,4 have a non nul constant term on the right. So they are rejected.
The case 1 : $\quad x=y\quad$ is convenient if we set $c_1=c_2$ . This is the expected choice.