求解常微分方程y'=(2*x-y-1)/(x-2*y+1)的通解,要求过程尽量详细,写出每一步?
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![求解常微分方程y'=(2*x-y-1)/(x-2*y+1)的通解,要求过程尽量详细,写出每一步?](/uploads/image/z/11691246-30-6.jpg?t=%E6%B1%82%E8%A7%A3%E5%B8%B8%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8By%27%3D%282%2Ax-y-1%29%2F%28x-2%2Ay%2B1%29%E7%9A%84%E9%80%9A%E8%A7%A3%2C%E8%A6%81%E6%B1%82%E8%BF%87%E7%A8%8B%E5%B0%BD%E9%87%8F%E8%AF%A6%E7%BB%86%2C%E5%86%99%E5%87%BA%E6%AF%8F%E4%B8%80%E6%AD%A5%3F)
求解常微分方程y'=(2*x-y-1)/(x-2*y+1)的通解,要求过程尽量详细,写出每一步?
求解常微分方程y'=(2*x-y-1)/(x-2*y+1)的通解,要求过程尽量详细,写出每一步?
求解常微分方程y'=(2*x-y-1)/(x-2*y+1)的通解,要求过程尽量详细,写出每一步?
此题最简单的解法:全微分法.
∵y'=(2x-y-1)/(x-2y+1) ==>(x-2y+1)dy-(2x-y-1)dx=0
==>(xdy+ydx)-2xdx-2ydy+dy+dx=0
==>d(xy)-d(x²)-d(y²)+dx+dy=0
==>d(xy-x²-y²+x+y)=0
==>xy-x²-y²+x+y=C (C是积分常数)
∴原方程的通解是xy-x²-y²+x+y=C (C是积分常数).
解题过程如上所示,可以参考常微分方程的书籍,上面有这类问题详细解法,祝好!