Question

(dy)/(dx)=2y+1+,y(0)=3;求此微分方程满足初始条件的特解

Answer 1

dy/dx =2y+1

∫dy/(2y+1) =∫ dx

(1/2)ln|2y+1| = x+C'

ln|2y+1| = 2x+2C'

2y+1 =Ce^(2x)

y= [-1+Ce^(2x)]/2

y(0) =3

3= [-1+C]/2

C=7

ie

y= [-1+7e^(2x)]/2

Answer 2

解:微分方程为dy/dx=2y+1，化为y'-2y=1，y'e⁻²ˣ-2e⁻²ˣy=e⁻²ˣ，(ye⁻²ˣ)'=e⁻²ˣ，ye⁻²ˣ=-0.5e⁻²ˣ+c(c为任意常数)，微分方程的通解为y=ce²ˣ-0.5

∵y(0)=3 ∴有3=c-0.5，得:c=3.5

微分方程的特解为y=3.5e²ˣ-0.

解微分方程

请参考