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# Answer to Question #25829 in Differential Equations for Matthew Lind

Question #25829
Solve the equation (1/x)u_x - (1/y)u_y = 2u on x&gt;0, y&gt;0 with the initial condition u(x,x) = x^2.
1
2013-03-11T09:45:32-0400
(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 =&gt; & 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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