Solutions to the PDE :$2V\frac{\partial I}{\partial V}+2W\frac{\partial I}{\partial W}=I$ While working on engineering problem, I came across this PDE: Let $c_1,c_2$ be two real numbers. Find a continuous function $I:\mathbb{R}_{\geq 0}\times\mathbb{R}_{\geq 0}\rightarrow \mathbb{R}$ such that 
1) For every $V,W$, we have: $$I(V,0)=c_1\sqrt{V},I(0,W)=c_2\sqrt{W}$$
2) $I|\mathbb{R}_{>0}\times\mathbb{R}_{>0}$ is a differentiable function that satisfies:
$$2V\frac{\partial I}{\partial V}+2W\frac{\partial I}{\partial W}=I$$
After some educated guesses, I know that for every $\lambda\in \mathbb{R}$, the follwoing is a solution:
$$I(V,W)=\lambda(c_1\sqrt{V}+c_2\sqrt{W})+(1-\lambda)(\sqrt{c_1^2V+c_2^2W})$$
Question: Are there any other solutions ?
It would also be great if someone can get all solutions.
Thank you

I will award a bounty worth 500 points for finding the set of all solutions.
 A: Introduce polar coordinates in the $(v,w)$-plane:
$$v=r\cos\phi,\quad w=r\sin\phi\qquad(0\leq r<\infty, \ 0\leq\phi\leq{\pi\over2})\ .$$
Then 
$$2r\ I_r=2r\>(I_vv_r+I_w w_r)=2r(I_v\cos\phi+I_w\sin\phi)=2v\>I_v+2w\>I_w=I\ .$$
In short: In the new variables $r$ and $\phi$ the original PDE appears as
$$2r\>I_r=I\ .$$
The solutions to this are the functions
$$I(r,\phi)=c(\phi)\sqrt{r}\ ,\tag{1}$$
where the function $$\phi\mapsto c(\phi)\qquad(0\leq\phi\leq{\pi\over2})$$ is arbitrary. In order to satisfy the given boundary conditions you have to impose the conditions $c(0)=c_1$, $\>c\bigl({\pi\over2}\bigr)=c_2$. In terms of $v$ and $w$ the functions $(1)$ appear as
$$I(v,w)=c\bigl({\rm arg}(v,w)\bigr)\>(v^2+w^2)^{1/4}\ .$$
It follows that there are many more solutions than you thought.
A: Given
$$
2 V \frac{\partial I}{\partial V} + 2 W \frac{\partial I}{\partial W} = I.\tag{1}
$$

The PDE (1) has the property that if $I$ is a solution and $a$ is a function of $V/W$, then $a(V/W) I$ is also a solution, as
$$
\begin{eqnarray}
\left[2 V \frac{\partial}{\partial V} + 2 W \frac{\partial}{\partial W} \right]
a(V/W) I
&=& a(V/W)
\left[2 V \frac{\partial}{\partial V} + 2 W \frac{\partial}{\partial W} \right] I\\
&& + I \left[2 V \frac{\partial}{\partial V} + 2 W \frac{\partial}{\partial W} \right]
a(V/W)\\
&& a(V/W) I + I \left[ \frac{2V}{W} - \frac{2VW}{W^2} \right] a'(V/W)\\
&=& a(V/W) I.\tag{2}
\end{eqnarray}
$$
So factors can be a function of $V/W$.

It is clear that PDE (1) is linear, but it has an even stronger property.
Let $I_1$ and $I_2$ be solutions of the PDE (1), then

$$
\Big( a_1 I_1^{\zeta} + a_2 I_2^{\zeta} \Big)^{1/\zeta}
$$
is also a solution. We find
$$
\begin{eqnarray}
\left[2 V \frac{\partial}{\partial V} + 2 W \frac{\partial}{\partial W} \right]
\Big( a_1 I_1^{\zeta} + a_2 I_2^{\zeta} \Big)^{1/\zeta}
&=& \Big( a_1 I_1^{2\zeta} + a_2 I_2^{2\zeta} \Big)^{1/\zeta-1} \times\\
&& \hspace{1em} \left\{ a_1 \left[2 V \frac{\partial I_1}{\partial V}
  + 2 W \frac{\partial I_1}{\partial W} \right] I_1^{\zeta-1} \right.\\
&& \hspace{2em} \left. +
a_2 \left[2 V \frac{\partial I_2}{\partial V}
  + 2 W \frac{\partial I_2}{\partial W} \right] I_2^{\zeta-1} \right\}\\
&=& \Big( a_1 I_1^{\zeta} + a_2 I_2^{\zeta} \Big)^{1/\zeta-1} \times
\Big( a_1 I_1^{\zeta} + a_2 I_2^{\zeta} \Big)\\
&=& \Big( a_1 I_1^{\zeta} + a_2 I_2^{\zeta} \Big).\tag{3}
\end{eqnarray}
$$

To solve the PDE (1), we consider the co-ordinates
$$
\begin{eqnarray}
p &=& \ln\Big(\sqrt[4]{V}\Big) + \ln\Big(\sqrt[4]{W}\Big),\\
q &=& \ln\Big(\sqrt[4]{V}\Big) - \ln\Big(\sqrt[4]{W}\Big),
\end{eqnarray}
$$
whence
$$
\begin{eqnarray}
\frac{\partial}{\partial V}
&=& \frac{\partial p}{\partial V} \frac{\partial}{\partial p}
  + \frac{\partial q}{\partial V} \frac{\partial}{\partial q}\\
&=& \frac{1}{4V} \frac{\partial}{\partial p}
  + \frac{1}{4V} \frac{\partial}{\partial q},\\
\frac{\partial}{\partial W}
&=& \frac{\partial p}{\partial W} \frac{\partial}{\partial p}
  + \frac{\partial q}{\partial W} \frac{\partial}{\partial q}\\
&=& \frac{1}{4W} \frac{\partial}{\partial p}
  - \frac{1}{4W} \frac{\partial}{\partial q},
\end{eqnarray}
$$
and therefore
$$
2 V \frac{\partial I}{\partial V} + 2 W \frac{\partial I}{\partial W}
=\frac{\partial I}{\partial p},
$$
so we obtain the equation
$$
\frac{\partial I}{\partial p} = I,
$$
and the solution is given by
$$
I = a(q) \exp\Big( p + b q \Big),
$$
where $a(q)$ is a function of $q$, i.e. constant with respect to $p$.
Filling in the co-ordinates $p$ and $q$, we get
$$
\begin{eqnarray}
I &=& a\left[ \ln\Big(\sqrt[4]{V}\Big) - \ln\Big(\sqrt[4]{W}\Big) \right] \times\\
&& \hspace{1em} \exp\left[ \ln\Big(\sqrt[4]{V}\Big) + \ln\Big(\sqrt[4]{W}\Big)
  + 4 \xi \ln\Big(\sqrt[4]{V}\Big) - 4 \xi \ln\Big(\sqrt[4]{W}\Big) \right]\\
&=& a\left[ \ln\Big(\sqrt[4]{V/W}\Big) \right]
  \sqrt[4]{V^{1+\xi} W^{1-\xi}}.\tag{4}
\end{eqnarray}
$$

However, due to (3) there are many solutions to build from (4).
Let us write
$$
I_k(V,W) = \sqrt[4]{V^{1+\xi_k} W^{1-\xi_k}},
$$
then the general solution is given by
$$
J_k(V,W) = \left( \sum a_{k_p}(V/W) I_{k_p}(U,V)^\zeta
  + \sum b_{k_p}(V/W)^\zeta J_{k_q}(U,V)^\zeta \right)^{1/\zeta},
$$
where we have used (2). This solution is a recursion.

The boundary condition
$$
\begin{eqnarray}
I(V,0) &=& c_1 \sqrt{V},\\
I(0,W) &=& c_2 \sqrt{W},
\end{eqnarray}
$$
is a restriction to the factors $a_k$ and $b_k$.
So we do find solutions like
$$
\begin{eqnarray}
I(V,W) &=& c_1 \sqrt{V} + c_2 \sqrt{W},\\
I(V,W) &=& \lambda c_1 \sqrt{V} + \lambda c_2 \sqrt{W}\\
&&  + (1-\lambda) \sqrt{ c_1^2 V + c_2^2 W },
\end{eqnarray}
$$
but also
$$
\begin{eqnarray}
I(V,W) &=& c_1 \sqrt{V} + c_2 \sqrt{W} + c_3 \sqrt[4]{VW},\\
I(V,W) &=& \lambda c_1 \sqrt{V} + \lambda c_2 \sqrt{W} + c_3 c\sqrt[8]{V^3W^5}\\
&&  + (1-\lambda) \sqrt{ c_1^2 V + c_2^2 W + c_3 \sqrt[4]{VW^3} }
  + c_4 \sqrt[8]{ V W ^3 + V^3 W }.
\end{eqnarray}
$$

The general solution is given by
$$
J_k(V,W) = \left( \sum a_{k_p}(V/W) I_{k_p}(U,V)^\zeta
  + \sum b_{k_p}(V/W)^\zeta J_{k_q}(U,V)^\zeta \right)^{1/\zeta},
$$
where
$$
I_k(V,W) = \sqrt[4]{V^{1+\xi_k} W^{1-\xi_k}},
$$
and the factors follow from the boundary condition.
