一阶微分方程题,5,和6,谢谢啦
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一阶微分方程题,5,和6,谢谢啦一阶微分方程题,5,和6,谢谢啦 一阶微分方程题,5,和6,谢谢啦5.∵xdy+ydx=(x^3y^2-x)dx==>d(xy)=x(x^2y^2-1)dx=
一阶微分方程题,5,和6,谢谢啦
一阶微分方程题,5,和6,谢谢啦
一阶微分方程题,5,和6,谢谢啦
5.∵xdy+ydx=(x^3y^2-x)dx
==>d(xy)=x(x^2y^2-1)dx
==>d(xy)/((xy)^2-1)=xdx
==>[1/(xy-1)-1/(xy+1)]d(xy)=2xdx
=ln│(xy-1)/(xy+1)│=x^2+ln│C│ (C是常数)
=(xy-1)/(xy+1)=Ce^(x^2)
=xy=(1+Ce^(x^2))/(1-Ce^(x^2))
==>y=(1+Ce^(x^2))/(x(1-Ce^(x^2)))
∴原方程的通解是y=(1+Ce^(x^2))/(x(1-Ce^(x^2))).
6.∵xdy/dx+x+sin(x+y)=0
==>x(dy/dx+1)+sin(x+y)=0
==>xd(x+y)/dx=-sin(x+y)
==>d(x+y)/sin(x+y)=-dx/x
==>ln│(cos(x+y)-1)/(cos(x+y)+1)│=-2ln││+ln│C│ (C是常数)
==>(cos(x+y)-1)/(cos(x+y)+1)=C/x^2
==>cos(x+y)=(x^2+C)/(x^2-C)
∴原方程的通解是cos(x+y)=(x^2+C)/(x^2-C)
∵当x=π/2时,y=0
代入通解,得C=-π^2/4
∴所求特解是cos(x+y)=(4x^2-π^2)/(4x^2+π^2).