y^n+3y'+2y=2x^2+x+1的通解 如果满意再加100分
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y^n+3y'+2y=2x^2+x+1的通解 如果满意再加100分
郭敦顒回答:
求y^n+3y'+2y=2x^2+x+1的通解
3y′ =2x^2+x+1-y^n-2y
∴y′=(2x²+x+1-y^n-2y) (1)
又y^n+3dy / dx +2y=2x^2+x+1
∴(y^n+2y)dx+3dy=(2x^2+x+1)dx
两边积分得,∫(y^n+2y)d x+∫3dy=∫(2x²+x+1)dx
∴∫(y^n+2y)d x=(2/3)x³+(1/2)x²+x-3y ( 2)
∴(y^n+2y)是导函数,其原函数是[(2/3)x³+(1/2)x²+x-3y]
等号左边∫(y^n+2y)d x=(dx/dy)∫(y^n+2y)dy
=(1/y′)[y^(n+1)/(n+1)+ y²+C],
代回(2)式得,(1/y′)[y^(n+1)/(n+1)+ y²+C]=(2/3)x³+(1/2)x²+x-3y
∴y′=[y^(n+1)/(n+1)+ y²+C]/[ (2/3)x³+(1/2)x²+x-3y] (3)
由(1)、(3得),
(2x²+x+1-y^n-2y)=[y^(n+1)/(n+1)+ y²+C]/[ (2/3)x³+(1/2)x²+x-3y]
∴方程的通解是:
[y^(n+1)/(n+1)+ y²+C] =(2x²+x+1-y^n-2y)[ (2/3)x³+(1/2)x²+x-3y]
此种类型的微分方程,我首次接触到,思考了半天方解出.这应是解微分方程的一种方法,已与积分方程有联系了,显然超出了常微分方程的范围.
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