How Do You Solve the PDE u_x + x u_y = u + x with Initial Condition u(x,0)=x^2?

[pde]

Solution of the good PDE ?

Find the solution of u of the equation

u_x + x u_y = u + x if u(x,0)=x^2,x>0.

 

[kosovtsov]

The general solution to your PDE is

u(x,y) = -1-x+e^x F(x^2-2y),

where F is an arbitrary function.

Your boundary condition leads to

F(t) = (1+t^{1/2}+t)exp(-t^{1/2}),

where t=x^2-2y.

 

[fluidistic]

The general solution to your PDE is

u(x,y) = -1-x+e^x F(x^2-2y),

where F is an arbitrary function.

Your boundary condition leads to

F(t) = (1+t^{1/2}+t)exp(-t^{1/2}),

where t=x^2-2y.

I'm not the original poster, but nice. May I know which method you used to obtain the general solution? Thank you!

 

[kosovtsov]

The main idea: If the homogeneous DE

\alpha(x,y)\frac{\partial s(x,y)}{\partial x}+\beta(x,y)\frac{\partial s(x,y)}{\partial x}=0

is solvable (it is sufficient to find any particular solution), then DE of the following type

\alpha(x,y)\frac{\partial p(x,y)}{\partial x}+\beta(x,y)\frac{\partial p(x,y)}{\partial x}=\xi(x,y)p(x,y)+f(x,y)

is solvable (at least formally) too. And general solution for p(x,y) can be found in principle from s(x,y) by means of only algebraic manipulations.

For given DE first of all we have to reduce the homogeneous part of initial DE

\frac{\partial u(x,y)}{\partial x}+x\frac{\partial u(x,y)}{\partial x}-u(x,y)=0

to (u=\exp(v))

\frac{\partial v(x,y)}{\partial x}+x\frac{\partial v(x,y)}{\partial x}-1=0.

The (particular) polynomial solution of homogeneous part of the last DE

\frac{\partial v(x,y)}{\partial x}+x\frac{\partial v(x,y)}{\partial x}=0.

is v(x,y)=x^2-2y, so its general solution is v(x,y)=F(x^2-2y).

To find general solutions for DEs on the way back it is quite enough here to seek particular solutions of nonhomogeneous DEs in forms v(x,y)=a(x),u(x,y)=b(x).