Question

高等数学求通解详细点

Answer 1

解：设y'=p，则y''=dp/dx=(dp/dy)(dy/dx)=pdp/dy
代入原方程，得pdp/dy=50sin³ycosy
==>pdp=50sin³ycosydy
==>pdp=50sin³yd(siny)
==>p²=25(siny)^4+C1 (C1是积分常数)
==>y'²=25(siny)^4+C1
∵y(1)=π/2，y'(1)=5
==>25=25+C1
==>C1=0
∴y'²=25(siny)^4
==>y'=5sin²y，或y'=-5sin²y (∵y'=-5sin²y不满足条件y'(1)=5，∴舍去)
==>dy/sin²y=5dx
==>-coty=5x+C2 (C2是积分常数)
∵y(1)=π/2 ==>0=5+C2
==>C2=-5
∴coty=-5(x-1)
故原方程在条件y(1)=π/2和y'(1)=5下的解是coty=5(1-x)。

Answer 2

微分方程的题目。令y'=P
y''=dp/dx=(dp/dy)(dy/dx)=p(dp/dy)
pdp=50sin^3ycosydy
p^2=25sin^4y+c1
根据题设p在y=π/2时值为5，故c1=0
故p=5sin^2y
1/sin^2ydy=5dx
-coty=5x+c2
因为y(1)=π/2，0=5+c2，故c2=-5
得coty=5-5x