常微分方程教程(丁同仁、李承治第二版)
2
(1)2y2 5(dy) 4dx
解：令y 由p
dy
2225cost,p sint
dy
25
25
sint,y 2cost,p
255
sint, x ,
a.当sint 0 dx y 2sint
d(2cost)
25
22sintdt
25
sint
dt x
dt C
( x C)] 2cos[( x C)]
b当sint 0 cost 1 y (2).x2 3(
dy2
) 1.dx
sht
et e tet e t
,cht ,sht
22
解：令x cht,p
dyshtshtshtsh2t
dy dx d(xht) dtdx333故
y
sh2t1
C
81
2t 2t(e e 2)d(2t) C
811t (sh2t ) C
24
22
(3).(dy) y x 0.dx
(e2t e 2t 4t) C
解：令x u,p v,y u2 v2,dy pdx2udu 2vdv vdu (2u v)du 2vdv
dv
du
2u v2v
uv
2
2 u
v2u齐次方程
令v t,u vt,
dv
t 1
2
2t 1
2t 1
tdv vdt
2dvv
dt
t vdv dt
vdv
2 2t2 t
2t 1
2t 12 2t2 t
dt
lnv
2t 1
C
2 2t2 t
2t 1
2dt C
2t t 2