Let , $a,b,c,d \in \Bbb R$ such that $c^2+d^2 \not =0$. Then the Cauchy problem $au_x+bu_y=e^{x+y}$ , $x,y\in \Bbb R$ with $u(x,y)=0 $ on $cx+dy=0$ has a unique solution if
(A) $ac+bd \not=0$.
(B) $ad-bc \not =0$.
(C) $ac-bd\not=0$
(D) $ad+bc \not=0$
Using Lagranges equations we get , $bx-ay=C_1$ and $u-\frac{a}{a+b}e^{x+y}=C_2$. Then the solution becomes $\displaystyle u(x,y)=\frac{a}{a+b}e^{x+y}+\phi(bx-ay)$. Then how I can proceed further to find out the answer.
On
We can write the auxillary equations as:
$$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{e^{x+y}}$$
On solving, we will obtain two equations : $$bx-ay=c_1$$ $$u-\frac{e^{x+y}}{a}=c_2$$
So, an implicit solution will be $$\phi\lgroup bx-ay,u-\frac{e^{x+y}}{a}\rgroup=0$$
We can now write $$u-\frac{e^{x+y}}{a}=F(bx-ay)$$ $$\Rightarrow u=\frac{e^{x+y}}{a} + F(bx-ay)$$
Using the initial conditions $u(x,y)=0$ on $cx+dy=0$ , we have $y=\frac{-cx}{d}$
Substituting, $$u=0=\frac{e^{x-\frac{cx}{d}}}{a} + F(bx+\frac{acx}{d})$$ Simply consider the second part of R. H. S, taking $x$ common and simplifying, we get $$F\lgroup x(\frac{bd+ac}{d})\rgroup$$
Thus, $bd + ac \neq 0$. Hence answer is option (A).
At $y=-\dfrac{cx}{d}, u=0$. So $\dfrac{a}{a+b}e^{(d-c)x}+\phi((bd-ac)x)=0$. The first term is always non zero. So $\phi$ must be chosen such that the prescribed condition holds. Equivalently this boils down to $bd-ac\neq 0$.