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求下面两个齐次方程的通解 y'=e^(y/x)+y/x xy'=(x^2+y^2)^1/2+y
人气:234 ℃ 时间:2020-05-23 19:27:54
解答
y' = e^(y/x) + y/x
(y = xL,y' = x*dL/dx + L)
x*dL/dx + L = e^L + L
[e^(-L)] dL = 1/x dx
-e^(-L) = ln|x| + C
e^(-L) = - ln(x) - C
L = - ln[- ln(x) - C]
y/x = - ln[- ln(x) - C]
y = - xln[C- ln(x)]
____________________________________________________
xy' = √(x² + y²) + y
(y = xL,y' = x*dL/dx + L)
x(x*dL/dx + L) = √(x² + x²L²) + xL
x(x*dL/dx + L) = x[√(1 + L²) + L]
x*dL/dx = √(1 + L²)
dL/√(1 + L²) = 1/x dx
ln|L + √(1 + L²)| = ln|x| + C
L + √(1 + L²) = e^(lnx + C) = Ce^x
y/x + √(1 + y²/x²) = Ce^x
y + √(x² + y²) = Cxe^x
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