Transformed pde but my answer doesn't match solution? $$\frac{d^2u}{dx^2} + \frac{d^2u}{dy^2} + \frac{du}{dx} + 2\frac{du}{dy} + 3u = 0$$
Let $u = ve^{ax + by}$ and find $a, b$ such that we can transform to the following equation
$$\frac{d^2v}{dx^2} + \frac{d^2v}{dy^2} + Av = 0$$
where $A$ is an arbitrary constant.
I used the chain rule to get the first and second derivatives in terms of $v$. 
$\frac{du}{dx} = \frac{du}{dv}\frac{dv}{dx}$
Then
$v_{xx} + v_{yy} + (a+1)v_x + (b+2)v_y + 3v = 0$
and letting $a = -1$ and $b = -2$ gives 
$v_{xx} + v_{yy} + Av = 0$
However I have a solution for this problem and it says, $a=-\frac{1}{2}$ and $b=-1$
Have I made a mistake or is the given solution incorrect?

$$
u_{xx} + u_{yy} + u_x+2u_y + 3u = 0
$$
use the sub
$$
u_x = v_x\mathrm{e}^{ax+by} + au\\
u_{xx} = v_{xx}\mathrm{e}^{ax+by} + 2av_x\mathrm{e}^{ax+by} + a^2u\\
u_y = v_y\mathrm{e}^{ax+by} + bu\\
u_{yy} = v_{yy}\mathrm{e}^{ax+by} + 2bv_y\mathrm{e}^{ax+by} + b^2u\\
$$
thus we get
$$
v_{xx}\mathrm{e}^{ax+by} + 2av_x\mathrm{e}^{ax+by} + a^2u + v_{yy}\mathrm{e}^{ax+by} + 2bv_y\mathrm{e}^{ax+by} + b^2u + v_x\mathrm{e}^{ax+by} + au + 2v_y\mathrm{e}^{ax+by} + 2bu + 3u = 0
$$
we get
$$
\mathrm{e}^{ax+by}\left[v_{xx} + 2av_x + v_{yy} + 2bv_y + v_x + 2v_y\right] + \left(a^2+b^2+a + 2b+3\right)v\mathrm{e}^{ax+by} = 0
$$
combining terms
$$
\mathrm{e}^{ax+by}\left[v_{xx} + v_{yy} + 2(b+1)v_y +(2a+1)v_x  + \left(a^2+b^2+a + 2b+3\right)\right]  = 0
$$
this means 
$$
2a+1 = 0 \implies a = -\frac{1}{2}\\
b+1 = 0 \implies b = -1.
$$
and to wrap it up
$$
a^2+b^2+a + 2b+3 = A = \left(-\frac{1}{2}\right)^2+(-1)^2 -\frac{1}{2}  -2 + 3=\frac{7}{4}
$$
