- #1

- 30

- 0

## Homework Statement

[tex]x*u_{x} + y*u_{y}= 1 + y^2[/tex]

u(x,1) = 1+ x; -infinity < x < +infinity

Solve this parametrically and in terms of x and y

## Homework Equations

We are supposed to solve this using the method of characteristics

## The Attempt at a Solution

My problem is that solving the equation parametrically and with x and y give two different solutions.

I will first try to solve this parametrically, using variables s and t.

let x = s along the initial line y = 1

**Is this assumption correct, or should the initial line be y = 0?**I have only solved problems where the initial line as been explicitly stated. I chose y = 1 because it made the equations work out, and because the initial condition was u(x,

**1**), so I am not sure if that is correct.

So,

dx/dt = x => x = s * e^t

dy/dt = y => y = Ce^t,

**since the initial line is y = 1, C= 1**; y = e^t

du/dt = 1 + y^2 => u = t + [tex]\frac{e^{2t}}{2}[/tex] + D

Now apply the initial condition:

u(x,1) = 1 + x = 0 + 1/2 + D => D = x + 1/2

So,

u = t + [tex]\frac{e^2t}{2}[/tex] + x + 1/2

since y = e^t that means t = ln|y| (|| is absolute value... I would also like to know if y is always positive so I can remove the absolute value.)

So the solution for u is:

u = ln|y| + (y^2)/2 + x + 1/2

That is solving the equation parametrically, now I am a supposed so solve the equation treating x as a parameter variable and s as the other parameter variable:

Rewrite the equation to:

[tex]u_{x}+ (y/x)*u_{y} = (1 + y^2)/x [/tex] (divided through by x)

so,

dy/dt = y/x => y = k(s)x (k is a function of s)

du/dt = (1+y^2)/x => u = (1+y^2)ln(x) + C(s) = (1+y^2)ln(x) + G(y/x)

Applying the initial condition:

u(x,1) = 1+x = 2ln(x) + G(1/x)

G(1/x) = 1+x - 2ln(x)

This is where I get confused:

First, the solutions I derived don't look at all like each other, and second for the later solution, I don't see how G(1/x) will tell me what G(y/x) is. I would appreciate any and all help, Thank You.