Goursat Problem Solution

Question

Show that the equation $u_{xy}+u_x=0$, $x\leq x_0$, $y\leq y_0$ has the solution $\alpha(x)e^{-y}+\beta(y)$ where $\alpha,\beta$ are arbitrary single variable functions.

I have understood that I can integrate wrt $x$ to get $u_y+u=\gamma(y)$ and I am stuck there.

Answer 1

$$u_y+u=\gamma(y)$$ This a first order linear ODE (no $dx$ in it) $\frac{du}{dy}+u(y)=\gamma(y)$ .

The solution of the associated homogeneous ODE, $\frac{du}{dy}+u=0$ is $u=c\:e^{-y}$

Replacing the constant $c$ by an unknown function $v(y)$ leads to $u=e^{-y}v(y)$ , $$u_y+u=e^{-y}(-v+v')+e^{-y}v=e^{-y}v'=\gamma(y)$$ $$\frac{dv}{dy}=e^y\gamma(y)$$ $v(y)=\int e^y\gamma(y)dy=f(y)+$constant relatively to $y$. $$v(x,y)=f(y)+\alpha(x)$$ $u(x,y)=e^{-y}v(x,y)=e^{-y}\left(f(y)+\alpha(x) \right)$ $$u(x,y)=\beta(y)+e^{-y}\alpha(x)$$ with $\beta(y)=e^{-y}f(y)$

Answer 2

$$u_{xy}+u_x=0$$ Substitute $u_x=p$ $$\frac {dp}{dy}+p=0$$ $$\int \frac {dp}{p}=-\int dy = -y+c(x)$$ $$\ln p=- y+c(x)$$ $$u_x=e^{- y}c(x)$$ $$\int du =\int e^{- y}c(x)dx$$ $$u(x,y) = e^{- y}\int c(x)dx$$ $$\boxed{u(x,y) = e^{- y}(K(x)+P (y))}$$