Method of characteristics - inhomogeneous pde

Question

The boundary conditions problem:

$2\frac{\partial u}{\partial x}-14\frac{\partial u}{\partial y}=3(x+y)$

Where the boundary conditions are:

$u(x,-6x)=Cos((2x)^2) \quad \forall x \in \mathbb{R}$

How would this be solved by method of characteristics?

My own attempt:

$a=2,b=-14,c=3(x+y)$

use that:

$\frac{dx}{ds}=a, \frac{dy}{ds}=b,\frac{du}{ds}=c \Rightarrow \frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}=ds$

$\frac{dx}{2}=\frac{dy}{-14}=\frac{du}{3(x+y)}=ds$

$\frac{dx}{dy}=\frac{a}{b}=\frac{2}{-14}$

$2dy=-14dx \Rightarrow \int2dy=\int-14dx \Rightarrow 2y=-14x+A$

$\Rightarrow A=2y+14x$

Then i've tried to isolate dx and dy and said dx-dy but that didn't go anywhere productive for me.

Answer 1

$$2\frac{\partial u}{\partial x}-14\frac{\partial u}{\partial y}=3(x+y)$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{2}=\frac{dy}{-14}=\frac{du}{3(x+y)}$$ A first characteristic equation comes from solving $\frac{dx}{2}=\frac{dy}{-14}$ : $$14x+2y=c_1$$ A second characteristic equation comes from solving $\frac{7x\,dx-y\,dy}{14}=\frac{du}{3}$ because :

$\frac{dx}{2}=\frac{dy}{-14}=\frac{7x\,dx-y_,dy}{14x+14y}=\frac{du}{3(x+y)}$ $$u-\frac{3}{28}(7x^2-y^2)=c_2$$ The general solution of the PDE on implicit form $c_2=F(c_1)$ is : $$\boxed{u(x,y)=\frac{3}{28}(7x^2-y^2)+F(14x+2y)}$$ $F$ is an arbitrary function.

I suppose that you can determine the function $F$ accordingly to the given condition.

Answer 2

Solve $$2u_x-14y_y=3(x+y),\quad (1)$$ $$u(x,-6x)=\cos^22x\qquad (2)$$ Steps of solution:

1 step:

General solution of $$2u_x-14y_y=0$$ is $$u=F(7x+y)$$ 2 step:

Particular solution of (1) is $$u=6x^2+\frac32xy$$

3 step:

Then general solution of (1) is $$u=6x^2+\frac32xy+F(7x+y)$$

4 step:

From (2) if $y=-6x$ we get $$F(x)=\cos^22x+3x^2$$

5 step:

Then solution of initial value problem (1), (2) is $$u=6x^2+\frac32xy+\cos^2(14x+2y)+3(7x+y)^2\\ =\cos^2(14x+2y)+153x^2+\frac{87}{2}xy+3y^2 $$