Solution to $u_x + y u_y = -u$ with $u(0, y) = \cos y$

Question

I am studying for my PDEs test and I want to make sure I can solve this type of equations.

I used the method of characteristics.

$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{b}{a} = y$$

Integrating I get $$C = ye^{-x}$$ where $C$ is a constant.

I can make the substitutions $$\xi = x$$ $$\eta = ye^{-x}$$ $$u(x,y) = w(\xi,\eta)$$

Substituting in the PDE, in terms of $w, \eta, \xi$ it becomes $$\frac{\mathrm{d}w}{\mathrm{d}\xi} + w = 0$$

I used the Integrating factor method to solve this ODE, which gives

$$w = f(\eta) e ^{-\xi}$$

If we substitute back to $u, x, y$ this becomes

$$w(\xi,\eta)=u(x,y)=f(ye^{-x})e^{-x}$$ The initial conditions give $$u(0,y)=\cos y =f(y)$$

Thus, the final solution is $$u(x,y) = \cos (ye^{-x})e^{-x}$$

Is this correct? Thanks in advance.

Answer 1

You can easily check whether it satisfies your PDE. With $u(x,y)=\cos(ye^{-x})e^{-x}$ you get $$\frac{\partial}{\partial x}u(x,y)+y\cdot\frac{\partial}{\partial y}u(x,y)= -e^{-x}\cos(ye^{-x})e^{-x}=-u(x,y)$$ and also $$u(0,y)=\cos(y).$$ Thus YES, your result is correct.

Answer 2

Thats correct

$$u_x + y u_y = -u$$ $$u_x + y u_y +u=0$$ $$u_x + \frac {\partial}{\partial_y}(yu)=0$$ $$\frac 1 y\frac {\partial}{\partial_x}(yu) + \frac {\partial}{\partial_y}(yu)=0$$ $$\frac {\partial}{\partial_x}(yu) +y \frac {\partial}{\partial_y}(yu)=0$$ We have that $uy=c_2$ $$\frac {dx}1=\frac {dy}y \implies c_1=\frac {e^x}y$$ $$uy=f(\frac {e^x}y)$$ $$u(x,y)=\frac 1 y f(\frac {e^x}y)$$ $$u(0,y)=\frac {f(1/y)}y \implies f(y)=\frac {\cos (1/y) }y$$ Therefore $$u(x,y)=e^{-x}\cos (ye^{-x})$$