求通解 方程：x^2ydx-(x^3+y^4)dy=0
1个回答
展开全部
这个方程要将 y 看成自变量.
x^2 * ydx - (x^3 + y^4)dy = 0 => x^2 * x ' * y - x^3 = y^4 =>
(x^2 * x ' * y - x^3) / y^4 = 1 => (x^2 * x ' * y^3 - x^3 * y^2) / y^6 = 1 =>
1/3 * ( x^3 / y^3 ) ' = 1 => ( x^3 / y^3 ) ' = 3 积分=> x^3 / y^3 = 3y + C
=> x^3 = y^3 * (3y + C) => x = y * (3y + C)^(1/3),C为任意常数.
x^2 * ydx - (x^3 + y^4)dy = 0 => x^2 * x ' * y - x^3 = y^4 =>
(x^2 * x ' * y - x^3) / y^4 = 1 => (x^2 * x ' * y^3 - x^3 * y^2) / y^6 = 1 =>
1/3 * ( x^3 / y^3 ) ' = 1 => ( x^3 / y^3 ) ' = 3 积分=> x^3 / y^3 = 3y + C
=> x^3 = y^3 * (3y + C) => x = y * (3y + C)^(1/3),C为任意常数.
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
