求微分方程(4y+3x)y′+y-2x=0的通解

原方程化为dy/dx=(2x-y)/(3x+4y)=(2-y/x)/(3+4y/x)
令y/x=u,则dy/dx=d(ux)/dx=u+du/dx
∴u+du/dx=(2-u)/(3+4u)
du/dx=(2-4u-4u^2)/(3+4u)
dx=[(3+4u)/(2-4u-4u^2)]du
两边积分即得我也算到了这一步怎么积分啊就这不会啊∫[(4u+3)/(2-4u-4u^2)]du=(-1/2)[∫[1/(4u^2+4u-2)]d(4u^2+4u-2)+2∫[1/(4u^2+4u-2)du]=(-1/2)[ln(4u^2+4u-2)+2∫[1/(4u^2+4u-2)du]其中∫[1/(4u^2+4u-2)du=∫1/[4(u+1/2)^2-6]d(u+1/2)令u+1/2=t得：∫1/(4t^2-6)dt=1/(4√6)*ln[(2t-√6)/(2t+√6)]+C代回去即可答案为（2y^2+2xy-x^2）^2=c((y+（1-根号3）x/2))/y+(1+根号3)x/2))^(1/根号3)啊有些乱啊