Charasteristic Method for PDE

Question

Hi i'm struggling a little with solutions of PDE. I have to solve the following

$$\begin{cases}\partial_xu +y^2\partial_yu = 2yu+y^2\\ u(0,y)=y \end{cases}$$

I want to use the method of characteristics so I write the system $$\begin{cases} \dot{x} = 1\\\dot{y} = y^2\\\dot{z} = 2yz+y^2\end{cases}$$ with $z(t)=u(x(t),y(t))$ and if this is a homogeneous PDE I can solve the first two row and I obtain that the characteristics curve are $$x +\frac{1}{y}=c$$ And the general solution of the homogeneous PDE is $$u(x,y) = F\bigg(x+\frac{1}{y}\bigg)$$ since this PDE isn't homogeneous I have to solve the last row and here comes the problems. I look at $\dot{z} = 2yz+y^2$ and I think its a ODE and I can solve it with the formula but then how can I use the characteristics to solve the entire equation, how can I find the general solution and how can I use the condition $u(0,y)=y^2$? I'm very confused on that part, thank you very much for the help.

Edit: 1) $u(x,y)=F\bigg(x+\frac{1}{y}\bigg)$ does not solve the homogeneous PDE but $\partial_xu +y^2\partial_yu =0$

Answer 1

$$\partial_xu +y^2\partial_yu = 2yu+y^2$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{1}=\frac{dy}{y^2}=\frac{du}{2yu+y^2}$$ A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{y^2}$ $$x+\frac{1}{y}=c_1$$ A second characteristic equation comes from solving $\frac{dy}{y^2}=\frac{du}{2yu+y^2}\quad\implies\quad \frac{du}{dy}=\frac{2u}{y}+1$ : $$\frac{u+y}{y^2}=c_2$$ The general solution of the PDE on the form of implicite equation $c_2=F(c_1)$ is : $$\frac{u+y}{y^2}=F\left(x+\frac{1}{y} \right)$$ where $F$ is an arbitrary function (to be determined according to the boundary condition). $$\boxed{u(x,y)=-y+y^2F\left(x+\frac{1}{y} \right)}$$ Condition :

$u(0,y)=y=-y+y^2F\left(0+\frac{1}{y} \right)\quad\implies\quad F\left(\frac{1}{y}\right)=\frac{2}{y}$

With $X=\frac{1}{y}$ the function $F(X)$ is determined : $$F(X)=2X$$ We put it into the above general solution where the argument is $X=x+\frac{1}{y}$.

$u(x,y)=-y+2y^2\left(x+\frac{1}{y}\right)$ $$u(x,y)=2y^2x+y$$

Answer 2

Since the PDE is linear and inhomogeneous, its general solution can be written as $u = v+w$, where $v$ is the solution to the homogeneous PDE, $$ v_x+y^2v_y=2yv, \tag{1} $$ and $w$ is a particular solution to the inhomogeneous equation. Let's first solve $(1)$. The characteristic equations are $$ dx=\frac{dy}{y^2}=\frac{dv}{2y}. \tag{2} $$ The solution to $dx=\frac{dy}{y^2}$ is $$ x+\frac{1}{y}=C_1, \tag{3} $$ whereas the solution to $\frac{dy}{y^2}=\frac{dv}{2y}$ is $$ v=C_2y^2, \tag{4} $$ hence the general solution to $(1)$ is $$ v(x,y)=y^2F\left(x+\frac{1}{y}\right), \tag{5} $$ where $F$ is an arbitrary differentiable function.

To find a particular solution $w$ to the inhomogeneous equation, let's make the ansatz $w=w(y)$. With this ansatz, $w_x=0$ and the PDE becomes a linear inhomogeneous ODE: $$ y^2w'(y)=2yw(y)+y^2. \tag{6} $$ One can easily check that $w(y)=-y$ is a particular solution to $(6)$. Combining it with $v(x,y)$, Eq. $(5)$, we conclude that the general solution to the inhomogeneous PDE is given by $$ u(x,y)=-y+y^2F\left(x+\frac{1}{y}\right). \tag{7} $$ To determine $F$, we use the condition $u(0,y)=y$: $$ -y+y^2F\left(\frac{1}{y}\right)=y \implies F\left(\frac{1}{y}\right)=\frac{2}{y} \implies F(t)=2t, \tag{8} $$ hence, finally, $$ u(x,y)=-y+2y^2\left(x+\frac{1}{y}\right)=2xy^2+y. \tag{9} $$