高数,微分方程求通解,求过程,谢谢😜
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(1) 微分方程即 2y ' = 1 - cos2(x-y+1) = 1 - cos2(y-x-1),
令 u = 2(y-x-1), 则 2y = u + 2x + 2, 2y' = u' + 2,
微分方程化为 u' + 2 = 1 - cosu,
du/dx = -1 - cosu = -2[cos(u/2)]^2,
[sec(u/2)]^2d(u/2) = -dx,
tan(u/2) = - x + C, 即 tan(y - x -1) = -x + C
(2) 微分方程即 y‘ + sinxcosy + cosxsiny = sinxcosy - cosxsiny
得 dy/dx = -2cosxsiny
dy/siny = -2cosxdx
ln(cscy - coty) = -2sinx + lnC
cscy-coty = Ce^(-2sinx)
令 u = 2(y-x-1), 则 2y = u + 2x + 2, 2y' = u' + 2,
微分方程化为 u' + 2 = 1 - cosu,
du/dx = -1 - cosu = -2[cos(u/2)]^2,
[sec(u/2)]^2d(u/2) = -dx,
tan(u/2) = - x + C, 即 tan(y - x -1) = -x + C
(2) 微分方程即 y‘ + sinxcosy + cosxsiny = sinxcosy - cosxsiny
得 dy/dx = -2cosxsiny
dy/siny = -2cosxdx
ln(cscy - coty) = -2sinx + lnC
cscy-coty = Ce^(-2sinx)
追问
看不懂
追答
(1) 先降幂,再换元, 再分离变量
(2) 先两边展开三角函数式,合并后分离变量
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