Solve PDE. Is this the right solution? $\frac{\partial u^{2}}{\partial x \partial y} + \frac{\partial u}{\partial y} + x + y + 1 = 0 $

Question

This is the equation:

$\frac{\partial u^{2}}{\partial x \partial y} + \frac{\partial u}{\partial y} + x + y + 1 = 0 $

I also have a solution but I don't know how it removed -1 in front of the arrow

Answer 1

$$u_{xy}+u_y+x+y+1=0$$ Integrate wrt $y$ : $$u_x+u+xy+\frac12 y^2+y=f(x)$$ $f(x)$ is an arbitrary function.

$u_x+u=-xy-\frac12 y^2-y+f(x)\quad$ is a first order linear ODE wrt $x$.

The solution of the associated homogeneous ODE $u_x+u=0$ is $u=\lambda\:e^{-x}$. Method of variation of the coefficient which becomes $\lambda(x,y)$ :

$u_x+u=(\lambda_xe^{-x}-\lambda e^{-x})+\lambda e^{-x}=-xy-\frac12 y^2-y+f(x)$

$\lambda_x=-xye^x-\frac12 y^2e^x-ye^x+f(x)e^x\quad$ that we integrate wrt $x$ :

$\lambda=(-xye^x+ye^x)-\frac12 y^2e^x-ye^x+F(x)+g(y)\quad$ where $F(x)$ and $g(y)$ are arbitrary functions.

$u=\lambda\:e^{-x}=-xy-\frac12 y^2-y+F(x)e^{-x}+g(y)e^{-x}$ $$\boxed{u(x,y)=-xy-\frac12 y^2-y+\varphi(x)+g(y)e^{-x}}$$ $\varphi(x)$ and $g(y)$ are arbitrary functions (to be determined according to some boundary conditions, if specified).