I was trying to solve the PDE:
$xu_{x} + 2yu_{y} = 3u$ with $u(x,y,0) = g(x,y)$
I thought of using method of characteristics
So the initial curve looks like $x = a, y=b , z = g(a,b)$
with
$\frac{dx}{ds} = x, \frac{dy}{ds} = 2y , \frac{dz}{ds} = 3z$
$x(0) =a, y(0) = b, z(0) = g(a,b)$
implying $x = ae^s , y = be^{2s} , z = g(a,b) e^{3s}$
So we have $xy = ab e^{3s}$
implying $e^{3s} = \frac{xy}{ab}$
so that $z = g(a,b) \frac{xy}{ab}$
but I think it is incorrect as i think that it doesnot satisfy the initial conditions.
Any help would be nice!
$$xu_x+2yu_y=3u\tag 1$$ $$\frac{dx}{x}=\frac{dy}{2y}=\frac{du}{3u}=ds$$ First characteristic equation from $\frac{dx}{x}=\frac{dy}{2y}$ : $$\frac{y}{x^2}=c_1$$ Second characteristic equation from $\frac{dx}{x}=\frac{du}{3u}$ : $$\frac{u}{x^3}=c_2$$ General solution of the PDE $(1)$ on the form of implicit equation $c_2=F(c_1)$ :
$\frac{u}{x^3}=F\left(\frac{y}{x^2}\right)$ $$u(x,y)=x^3F\left(\frac{y}{x^2}\right)$$ $F$ is an arbitrary function, to be determined according to some boundary condition.
The wording of the question makes appear a dummy variable $t$ . Thus the general solution of equation $(1)$ is : $$u(x,y,t)=x^3\Phi\left(\frac{y}{x^2}\:,\:t\right) \tag 2$$ $\Phi$ is an arbitrary function of two variables, to be determined according to some boundary and initial conditions.
The specified initial condition $u(x,y,0)=g(x,y)$ implies only one equation : $$g(x,y)=x^3\Phi\left(\frac{y}{x^2}\:,\:0\right)$$ which is not sufficient for the function $\Phi$ to be determined.