Case $1$: $a=b=0$ and $c\neq0$
Then $c\dfrac{\partial^2v}{\partial y^2}=10x^2+9x+6$
$\dfrac{\partial^2v}{\partial y^2}=\dfrac{10x^2+9x+6}{c}$
$\dfrac{\partial v}{\partial y}=\dfrac{(10x^2+9x+6)y}{c}+C_1(x)$
$v(x,y)=\dfrac{(10x^2+9x+6)y^2}{2c}+C_1(x)y+C_2(x)$
$v(x,0)=0$ :
$C_2(x)=0$
$\therefore v(x,y)=\dfrac{(10x^2+9x+6)y^2}{2c}+C_1(x)y$
$v(0,y)=0$ :
$\dfrac{6y^2}{2c}+C_1(0)y=0$
$C_1(0)=-\dfrac{3y}{c}$
$\therefore v(x,y)=\dfrac{(10x^2+9x+6)y^2}{2c}+C_1(x)y$ , where $C_1(x)$ is any function satisfy $C_1(0)=-\dfrac{3y}{c}$
Case $2$: $a=c=0$ and $b\neq0$
Then $bx\dfrac{\partial v}{\partial x}=10x^2+9x+6$
$\dfrac{\partial v}{\partial x}=\dfrac{10x}{b}+\dfrac{9}{b}+\dfrac{6}{bx}$
$v(x,y)=\dfrac{5x^2}{b}+\dfrac{9x}{b}+\dfrac{6\ln x}{b}+C(y)$
But $x$ cannot substitute $0$ as it will becomes undefined
$\therefore$ There is no solution when we force to satisfy the I.C. $v(0,y)=0$
Case $3$: $a\neq0$ and $b\neq0$ and $c=0$
Then $ax^2\dfrac{\partial^2v}{\partial x^2}+bx\dfrac{\partial v}{\partial x}=10x^2+9x+6$
$\dfrac{\partial^2v}{\partial x^2}+\dfrac{b}{ax}\dfrac{\partial v}{\partial x}=\dfrac{10}{a}+\dfrac{9}{ax}+\dfrac{6}{ax^2}$
I.F. $=e^{\int\frac{b}{ax}dx}=e^{\frac{b}{a}\ln x}=e^{\ln x^{\frac{b}{a}}}=x^{\frac{b}{a}}$
$\therefore\dfrac{\partial}{\partial x}\left(x^{\frac{b}{a}}\dfrac{\partial v}{\partial x}\right)=\dfrac{10x^{\frac{b}{a}}}{a}+\dfrac{9x^{\frac{b}{a}-1}}{a}+\dfrac{6x^{\frac{b}{a}-2}}{a}$
$\begin{cases}x^{\frac{b}{a}}\dfrac{\partial v}{\partial x}=\dfrac{10x^{\frac{b}{a}+1}}{a+b}+\dfrac{9x^{\frac{b}{a}}}{b}+\dfrac{6x^{\frac{b}{a}-1}}{b-a}+C_1(y)&\text{when}\dfrac{b}{a}\neq-1,0,1\\\dfrac{1}{x}\dfrac{\partial v}{\partial x}=\dfrac{10\ln x}{a}-\dfrac{9}{ax}-\dfrac{3}{ax^2}+C_1(y)&\text{when}\dfrac{b}{a}=-1\\\dfrac{\partial v}{\partial x}=\dfrac{10x}{a}+\dfrac{9\ln x}{a}-\dfrac{6}{ax}+C_1(y)&\text{when}\dfrac{b}{a}=0\\x\dfrac{\partial v}{\partial x}=\dfrac{5x^2}{a}+\dfrac{9x}{a}+\dfrac{6\ln x}{a}+C_1(y)&\text{when}\dfrac{b}{a}=1\end{cases}$
$\dfrac{\partial v}{\partial x}=\begin{cases}\dfrac{10x}{a+b}+\dfrac{9}{b}+\dfrac{6}{(b-a)x}+C_1(y)x^{-\frac{b}{a}}&\text{when}\dfrac{b}{a}\neq-1,0,1\\\dfrac{10x\ln x}{a}-\dfrac{9}{a}-\dfrac{3}{ax}+C_1(y)x&\text{when}\dfrac{b}{a}=-1\\\dfrac{10x}{a}+\dfrac{9\ln x}{a}-\dfrac{6}{ax}+C_1(y)&\text{when}\dfrac{b}{a}=0\\\dfrac{5x}{a}+\dfrac{9}{a}+\dfrac{6\ln x}{ax}+\dfrac{C_1(y)}{x}&\text{when}\dfrac{b}{a}=1\end{cases}$
$v(x,y)=\begin{cases}\dfrac{5x^2}{a+b}+\dfrac{9x}{b}+\dfrac{6\ln x}{b-a}+\dfrac{C_1(y)ax^{1-\frac{b}{a}}}{a-b}+C_2(y)&\text{when}\dfrac{b}{a}\neq-1,0,1\\\dfrac{5x^2(2\ln x-1)}{2a}-\dfrac{9x}{a}-\dfrac{3\ln x}{a}+\dfrac{C_1(y)x^2}{2}+C_2(y)&\text{when}\dfrac{b}{a}=-1\\\dfrac{5x^2}{a}+\dfrac{9x(\ln x-1)}{a}-\dfrac{6\ln x}{a}+C_1(y)x+C_2(y)&\text{when}\dfrac{b}{a}=0\\\dfrac{5x^2}{2a}+\dfrac{9x}{a}+\dfrac{3(\ln x)^2}{a}+C_1(y)\ln x+C_2(y)&\text{when}\dfrac{b}{a}=1\end{cases}$
But $x$ cannot substitute $0$ as it will becomes undefined
$\therefore$ There is no solution when we force to satisfy the I.C. $v(0,y)=0$
Case $4$: $a=0$ and $b\neq0$ and $c\neq0$
Then $bx\dfrac{\partial v}{\partial x}+c\dfrac{\partial^2v}{\partial y^2}=10x^2+9x+6$
It is possible that the subsititution $v(x,y)=v_c(x,y)+v_p(x,y)$ can make the inhomogeneous linear PDE becomes homogeneous linear PDE if $v_p(x,y)$ can be found.
For this question, the form of $v_p(x,y)$ is not difficult to guess, just the particular solution of the ODE $bx\dfrac{dv}{dx}=10x^2+9x+6$
But with reference to case $2$ , this ODE is no solution when we force to satisfy the I.C. $v(0,y)=0$
$\therefore$ There is no solution when we force to satisfy the I.C. $v(0,y)=0$
Case $5$: $a\neq0$ and $b\neq0$ and $c\neq0$
Then $ax^2\dfrac{\partial^2v}{\partial x^2}+bx\dfrac{\partial v}{\partial x}+c\dfrac{\partial^2v}{\partial y^2}=10x^2+9x+6$
It is possible that the subsititution $v(x,y)=v_c(x,y)+v_p(x,y)$ can make the inhomogeneous linear PDE becomes homogeneous linear PDE if $v_p(x,y)$ can be found.
For this question, the form of $v_p(x,y)$ is not difficult to guess, just the particular solution of the ODE $ax^2\dfrac{d^2v}{dx^2}+bx\dfrac{dv}{dx}=10x^2+9x+6$
But with reference to case $3$ , this ODE is no solution when we force to satisfy the I.C. $v(0,y)=0$
$\therefore$ There is no solution when we force to satisfy the I.C. $v(0,y)=0$