题目
【题目】求微分方程 (3x^2+2xy-y^2)dx+(x^2-2xy)dy=0的通解
【题目】求微分方程 (3x^2+2xy-y^2)dx+(x^2-2xy)dy=0的通解
题目解答
答案
【解析】令y=m，则 u=y/x ，且 (dy)/(dx)=u+x(du)/(dx)由 (3x^2+2xy-y^2)dx+(x^2-2xy)dy=0 可得(dy)/(dx)=-(3x^2+2xy-y^2)/(x^2-2xy)= -(u^2-2u-3)/(2u-1) 所以x(du)/(dx)=(dy)/(dx)-u=-(3(u^2-u-1))/(2u-1)利用分离变量可得,(2u-1)/(u^2-u-1)du=-3/xdx u2-u-1两边积分可得故 u^2-u-1=C/(x^3)将 u=y/x 入，可得y^2-xy-x^2=C/x
解析
步骤 1：引入变量
令 \(y = mx\)，其中 \(m\) 是 \(x\) 的函数，即 \(m = \frac{y}{x}\)。因此，\(y = mx\)，且 \(\frac{dy}{dx} = m + x\frac{dm}{dx}\)。
步骤 2：代入原方程
将 \(y = mx\) 和 \(\frac{dy}{dx} = m + x\frac{dm}{dx}\) 代入原方程 \((3x^2 + 2xy - y^2)dx + (x^2 - 2xy)dy = 0\)，得到
\[
(3x^2 + 2x(mx) - (mx)^2)dx + (x^2 - 2x(mx))(m + x\frac{dm}{dx})dx = 0
\]
化简得
\[
(3x^2 + 2mx^2 - m^2x^2)dx + (x^2 - 2mx^2)(m + x\frac{dm}{dx})dx = 0
\]
\[
x^2(3 + 2m - m^2)dx + x^2(1 - 2m)(m + x\frac{dm}{dx})dx = 0
\]
\[
x^2(3 + 2m - m^2) + x^2(1 - 2m)(m + x\frac{dm}{dx}) = 0
\]
\[
x^2(3 + 2m - m^2) + x^2(1 - 2m)m + x^3(1 - 2m)\frac{dm}{dx} = 0
\]
\[
x^2(3 + 2m - m^2 + m - 2m^2) + x^3(1 - 2m)\frac{dm}{dx} = 0
\]
\[
x^2(3 + 3m - 3m^2) + x^3(1 - 2m)\frac{dm}{dx} = 0
\]
\[
x^2(3(1 + m - m^2)) + x^3(1 - 2m)\frac{dm}{dx} = 0
\]
\[
3(1 + m - m^2) + x(1 - 2m)\frac{dm}{dx} = 0
\]
\[
x(1 - 2m)\frac{dm}{dx} = -3(1 + m - m^2)
\]
\[
\frac{dm}{dx} = \frac{-3(1 + m - m^2)}{x(1 - 2m)}
\]
步骤 3：分离变量并积分
分离变量得
\[
\frac{1 - 2m}{1 + m - m^2}dm = -\frac{3}{x}dx
\]
两边积分得
\[
\int \frac{1 - 2m}{1 + m - m^2}dm = -3\int \frac{1}{x}dx
\]
\[
\int \frac{1 - 2m}{1 + m - m^2}dm = -3\ln|x| + C
\]
\[
\int \frac{1 - 2m}{1 + m - m^2}dm = -3\ln|x| + C
\]
\[
\ln|1 + m - m^2| = -3\ln|x| + C
\]
\[
1 + m - m^2 = Cx^{-3}
\]
\[
1 + \frac{y}{x} - \frac{y^2}{x^2} = Cx^{-3}
\]
\[
x^2 + xy - y^2 = C
\]
令 \(y = mx\)，其中 \(m\) 是 \(x\) 的函数，即 \(m = \frac{y}{x}\)。因此，\(y = mx\)，且 \(\frac{dy}{dx} = m + x\frac{dm}{dx}\)。
步骤 2：代入原方程
将 \(y = mx\) 和 \(\frac{dy}{dx} = m + x\frac{dm}{dx}\) 代入原方程 \((3x^2 + 2xy - y^2)dx + (x^2 - 2xy)dy = 0\)，得到
\[
(3x^2 + 2x(mx) - (mx)^2)dx + (x^2 - 2x(mx))(m + x\frac{dm}{dx})dx = 0
\]
化简得
\[
(3x^2 + 2mx^2 - m^2x^2)dx + (x^2 - 2mx^2)(m + x\frac{dm}{dx})dx = 0
\]
\[
x^2(3 + 2m - m^2)dx + x^2(1 - 2m)(m + x\frac{dm}{dx})dx = 0
\]
\[
x^2(3 + 2m - m^2) + x^2(1 - 2m)(m + x\frac{dm}{dx}) = 0
\]
\[
x^2(3 + 2m - m^2) + x^2(1 - 2m)m + x^3(1 - 2m)\frac{dm}{dx} = 0
\]
\[
x^2(3 + 2m - m^2 + m - 2m^2) + x^3(1 - 2m)\frac{dm}{dx} = 0
\]
\[
x^2(3 + 3m - 3m^2) + x^3(1 - 2m)\frac{dm}{dx} = 0
\]
\[
x^2(3(1 + m - m^2)) + x^3(1 - 2m)\frac{dm}{dx} = 0
\]
\[
3(1 + m - m^2) + x(1 - 2m)\frac{dm}{dx} = 0
\]
\[
x(1 - 2m)\frac{dm}{dx} = -3(1 + m - m^2)
\]
\[
\frac{dm}{dx} = \frac{-3(1 + m - m^2)}{x(1 - 2m)}
\]
步骤 3：分离变量并积分
分离变量得
\[
\frac{1 - 2m}{1 + m - m^2}dm = -\frac{3}{x}dx
\]
两边积分得
\[
\int \frac{1 - 2m}{1 + m - m^2}dm = -3\int \frac{1}{x}dx
\]
\[
\int \frac{1 - 2m}{1 + m - m^2}dm = -3\ln|x| + C
\]
\[
\int \frac{1 - 2m}{1 + m - m^2}dm = -3\ln|x| + C
\]
\[
\ln|1 + m - m^2| = -3\ln|x| + C
\]
\[
1 + m - m^2 = Cx^{-3}
\]
\[
1 + \frac{y}{x} - \frac{y^2}{x^2} = Cx^{-3}
\]
\[
x^2 + xy - y^2 = C
\]