I have to solve 3 * du/dx - 4*du/dy + u = 1 + x/3
with initial condition u(0,y)= y ^2 using the method of characteristics
The above derivatives are partial derivatives
My working until now:
We have dx/ds = 3 =>x=3s+x(0) => s=(x-x(0))/3
dy/ds = -4 => y= -4s + y(0) => y= -4(x-x(0))/3 + y(0) {using that s=(x-x(0))/3}
du/ds = 1+ x/3 - u
But since s=(x-x(0))/3 => du/ds = 1+s+x(0)/3 - u
The above were our characteristic equations.
From our initial conditions now, x(0)=0 so x=3s
Also y= -4(x-x(0))/3 + y(0) from above becomes y+ 4/3 * x = y(0)
Furthermore du/ds = 1+s - u {since x(0)=0}
Solving using the integrating factor yields that u(s) = s + c*e^(-s), where c is an arbitrary constant!
Moreover u(s=0) = u(0) = y(0) ^2 {using the initial condition u(0,y) = y^2}
Finally, since x = 3s from above, we deduce u(x/3) = x/3 + (y + 4/3 * x)^2 * e^(-x/3)
Then I don't know how to proceed to find u(x,y)
May someone review my work and help me please?
Thanks!!!