> 数学 >
微积分微分方程问题1
求微分方程xy dy/dx = x^2+Y^2满足初始条件的Y|x=e =2e的特解
人气:390 ℃ 时间:2020-06-03 10:31:39
解答
xydy/dx=x^2+y^2
xydy=x^2dx+y^2dx
xdy^2=2x^2dx+2y^2dx
y^2=ux
dy^2=udx+xdu
x(udx+xdu)=2x^2dx+2uxdx
udx+xdu=2xdx+2udx
xdu-udx=2xdx
du/x-udx/x^2=2dx/x
d(u/x)=2dlnx
u/x=C+2lnx
u=Cx+2xlnx
y^2/x=Cx+2xlnx
通解
y^2=Cx^2+2x^2lnx
y|x=e =2e
4e^2=C4e^2+8e^2
C= -1
特解y^2=2x^2lnx-x^2
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版