Question #25829

Solve the equation (1/x)u_x - (1/y)u_y = 2u on x>0, y>0 with the initial condition u(x,x) = x^2.

Expert's answer

(1/x)u_x - (1/y)u_y = 2u, u(x,0) = x^2.

Method of characteristics:

dx/(1/x) = -dy/(1/y) = du/2u

xdx = ydy = du/2u

1) xdx = -ydy

& Integrate:

x^2/2 = -y^2/2 + C1/2

C1 = x^2 + y^2

2) -ydy = du/2u

Integrate:

-y^2/2 = (1/2)ln(u) - C2/2

C2 = y^2 + ln(u)

Therefore, the complete integral of the equation:

F ( x^2 + y^2,& y^2 + ln(u) ) = 0

or in another form:

ln(u) =& f (x^2 + y^2) - y^2

u = exp[ f (x^2 + y^2) -y^2]

where f - some differentiable function

The initial condition u(x,0) = x^2

u(x,0) = exp[ f (x^2 - 0) -0] = exp[ f(x^2) ] = x^2 => & f(x^2)=ln(x^2)

u =& (x^2 + y^2)exp[-y^2]

Answer: u =& (x^2 + y^2)exp[-y^2]

Method of characteristics:

dx/(1/x) = -dy/(1/y) = du/2u

xdx = ydy = du/2u

1) xdx = -ydy

& Integrate:

x^2/2 = -y^2/2 + C1/2

C1 = x^2 + y^2

2) -ydy = du/2u

Integrate:

-y^2/2 = (1/2)ln(u) - C2/2

C2 = y^2 + ln(u)

Therefore, the complete integral of the equation:

F ( x^2 + y^2,& y^2 + ln(u) ) = 0

or in another form:

ln(u) =& f (x^2 + y^2) - y^2

u = exp[ f (x^2 + y^2) -y^2]

where f - some differentiable function

The initial condition u(x,0) = x^2

u(x,0) = exp[ f (x^2 - 0) -0] = exp[ f(x^2) ] = x^2 => & f(x^2)=ln(x^2)

u =& (x^2 + y^2)exp[-y^2]

Answer: u =& (x^2 + y^2)exp[-y^2]

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