Number of solutions of The Cauchy Problem $\begin{cases} 2u_x+3u_y=5\\u=1\ \text{on the line} \ 3x-2y=0\end{cases}$

Question

The Cauchy Problem $$\begin{cases} 2u_x+3u_y=5\\u=1\ \text{on the line} \ 3x-2y=0\end{cases}$$ has
$(a) $exactly one solution
$(b) $ exactly two solutions
$(c) $ infinitely many solutions
$(d)$no solution
$\color{blue}{Try:} $
The auxiliary equation is $\frac{dx}2=\frac{dy}3=\frac{du}5$
From the first two fraction we get $3x-2y=c_1$ and from the last two fraction we get $5y-3u=c_2$
So, general solution is $3x-2y=\phi(5x-2u)$
Now, by given condition we have
$0=\phi(5x-2)$
I'm unable to conclude from here.
Also If instead of doing the last two lines if we put $3x-2y=0$ in $3x-2y=c_1$ then we get $c_1=0$. What to do next?
If there is any alternative approach, please teach me that. Please help me. Thanks in advance.

Answer 1

Your Cauchy problem has no solution, indeed

if there were a solution $\;u(x,y)\;,\;$ then $\;f(x)=u\left(x,\dfrac{3x}2\right)=1\;$ for all $\;x\in\mathbb{R}\;.$

Consequently,

$f’(x)=u_x+\dfrac32u_y=0\;$ for all $\;x\in\mathbb{R}\;,\;$ that is

$2u_x+3u_y=0\;$ which is a contradiction.