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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