As Herebrij and David pointed out; if one of the is zero, say $b=0$, then we have a unique solution.
Using Method of Characteristics, letting $x=x(t),y=y(t)$ and hence $u=u(t)$
$$\frac{du}{dt}=\frac{\partial u}{\partial x}\frac{dx}{dt}+\frac{\partial u}{\partial y}\frac{dy}{dt}$$
Comparing to $a\frac{\partial u}{\partial x}+b\frac{\partial u}{\partial x}=1$, we get
$$\frac{dx}{dt}=a,\frac{dy}{dt}=b, \frac{du}{dt}=1$$
$\frac{dx}{dt}=a\implies x=at\implies dt=\frac{dx}a$
$\frac{dy}{dt}=b\implies a\frac{dy}{dx}=b\implies ay-bx=y_0$
$\frac{du}{dt}=1\implies a\frac{du}{dx}=1\implies u(x,y)={x+F(y_0)\over a}={x+F(ay-bx)\over a}$
Now, $(\frac 1ax,0)$ lies on $ax+by=1$, so
$$u\left(\frac 1a x,0\right)=\frac 1a\left[\frac xa+F\left(-b\frac xa\right)\right]=\frac xa\implies F\left(-b\frac xa\right)=\frac xa\left[1-\frac 1a\right]$$
Letting $t=-b\frac xa$, we get
$$F(t)=\frac t{-b}\left[1-\frac 1a\right]$$
Substituting and simplifying, $u(x,y)=\frac{y(1-a)}{b}+x$
Which is unique for given $(a,b)$. Hence option 3.
The integration you performed is not correct as $x$ has dependence on $y$ (in fact, the first equation tell us how they are related).
Nevertheless, we can integrate the equations. Further, we get an explicit expression for $u$. I take as starting point these equations:
$$\frac{dx}{\cos(ky)}=\frac{dy}{1}=\frac{du}{ax^2}$$
From them:
$dx=dy\cos(ky)$ and $\dfrac{ax^2dx}{\cos(ky)}=du$ From the first one,
$x+c_1=\dfrac{1}{k}\sin(ky)$ or $c_1=\dfrac{1}{k}\sin(ky)-x$
From here, we can write the cosine as function of $x$ to integrate the second one:
$\sqrt{1-k^2(x+c_1)^2}=\cos(ky)$
Substituting into the second one
$\dfrac{ax^2dx}{\sqrt{1-k^2(x+c_1)^2}}=du$
And integrating:
$$u+c_2=\dfrac{a\left((2k^2c_1^2+1)\arcsin(k(x+c_1))+k(3c_1-x)\sqrt{1-k^2(x+c_1)^2}\right)}{2k^3}$$
Eliminating $c_1$
$$u+c_2=\dfrac{a\left(\left(2k^2\left(\dfrac{1}{k}\sin(ky)-x\right)^2+1\right)y+\left(\dfrac{3}{k}\sin(ky)-4x\right)\cos(ky)\right)}{2k^2}$$
At last, considering that $c_2=f(c_1)$ with $f$ a single argument differentiable function, the general solution is:
$$u(x,y)=\dfrac{a\left(\left(2k^2\left(\dfrac{1}{k}\sin(ky)-x\right)^2+1\right)y+\left(\dfrac{3}{k}\sin(ky)-4x\right)\cos(ky)\right)}{2k^2}-f\left(\dfrac{1}{k}\sin(ky)-x\right)$$
Now, the boundary conditions $u(x,0)=0$ impose some restriction for $f$:
$$u(x,0)=\dfrac{-2ax}{k^2}-f(-x)=0$$
So $f(x)=\dfrac{2ax}{k^2}$ and
$$u(x,y)=\dfrac{a\left(\left(2k^2\left(\dfrac{1}{k}\sin(ky)-x\right)^2+1\right)y+\left(\dfrac{3}{k}\sin(ky)-4x\right)\cos(ky)\right)}{2k^2}-\dfrac{2a}{k^2}\left(\dfrac{1}{k}\sin(ky)-x\right)$$
$$u(x,y)=\dfrac{a}{2k^2}\left(\left(2k^2\left(\dfrac{1}{k}\sin(ky)-x\right)^2+1\right)y+\left(\dfrac{3}{k}\sin(ky)-4x\right)\cos(ky)-\left(\dfrac{1}{k}\sin(ky)-x\right)4\right)$$
Best Answer
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).