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標題:
發問:
solve y'=(x^2+y^2)/(2xy), y(1)=-2
最佳解答:
Homogeneous ODE. Let u = y/x. y = xu y' = u + xu' y' = (x2 + y2)/(2xy) u + xu' = (x2 + x2u2)/(2x2u) u + xu' = (1 + u2)/(2u) xu' = (1 + u2)/(2u) - u x(du/dx) = (1 - u2)/(2u) 2u/(1 - u2) du = dx/x 1/(1 - u2) d(u2) = dx/x -ln|1 - u2| = ln|x| + C -ln|1 - y2/x2| = ln|x| + C y(1) = -2 -ln|1 - 4/1| = ln|1| + C C = -ln 3 -ln|1 - y2/x2| = ln|x| - ln 3 ln|x2/(x2 - y2)| = ln|x/3| |x2/(x2 - y2)| = |x/3| |x2 - y2| = 3|x| x2 - y2 = ±3x y2 = x2 ± 3x y = -√(x2 + 3x) based on the initial condition.
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initial value problem?發問:
solve y'=(x^2+y^2)/(2xy), y(1)=-2
最佳解答:
Homogeneous ODE. Let u = y/x. y = xu y' = u + xu' y' = (x2 + y2)/(2xy) u + xu' = (x2 + x2u2)/(2x2u) u + xu' = (1 + u2)/(2u) xu' = (1 + u2)/(2u) - u x(du/dx) = (1 - u2)/(2u) 2u/(1 - u2) du = dx/x 1/(1 - u2) d(u2) = dx/x -ln|1 - u2| = ln|x| + C -ln|1 - y2/x2| = ln|x| + C y(1) = -2 -ln|1 - 4/1| = ln|1| + C C = -ln 3 -ln|1 - y2/x2| = ln|x| - ln 3 ln|x2/(x2 - y2)| = ln|x/3| |x2/(x2 - y2)| = |x/3| |x2 - y2| = 3|x| x2 - y2 = ±3x y2 = x2 ± 3x y = -√(x2 + 3x) based on the initial condition.
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