The linear problem is given as $$x\frac{\text{$\delta $u}\backslash }{\text{$\delta $x}}\text{+y}\frac{\text{$\delta $u}\backslash }{\text{$\delta $y}}\text{=(x+y)u}$$ with $u = 1$ on $x=1$ with $1<y<2$
I worked the answer to be $$\log[x]=x + y \log(x)$$ Hence, $$u = x e^{x+y}$$ but in the text, $$\log[u] = \left(1+\frac{y}{x}\right)(x-1)$$
Which is very strange.
Did I went wrong somewhere?
the characteristics are given by $$\frac{dx}{dt} = x, \frac{dy}{dt} = y, x(0) = 1, y(0) = b $$ which gives $$x = e^t, y = be^t, b = \frac yx, t=\ln x.$$ along the characteristics we have $$\frac{du}{dt} = u(1+b)e^t, u = 1, t=0 \to \int_1^u\frac{du}{u} = (1+b)\int_0^t e^t\, dt$$ integrating we get $$\ln u=(1+b)(e^t-1). u(x, y) = e^{(1+b)}e^{(e^t)}=e^{1+y/x}e^{x-1} =e^{x + y/x}$$
so we have $$u = e^{x+ y/x} $$