Question

求微分方程的通解？？

Answer 1

dS=(-S*cost+1/2 * sin2t )*dt
= (-S*cost+sint*cost)*dt
=(-S+sint) *cost*dt
=(-S+sint) dsint
令sint=u
dS=(-S+u)du
dS/du=-S+u
S'+S=u

求 S‘+S=0通解
得到S0=Ce^(-u)

求S'+S=u 特解
得到S1=u-1

则完全通解是
S= u-1+Ce^(-u)
带回 u=sint

得到通解： S= sint-1 + Ce^(-sint)

Answer 2

Answer 3

t = ((1/4)*sin((x-1)*t)/(x-1)-(1/4)*sin((x+1)*t)/(x+1)+C)/sin(t)

Answer 4

Answer 5

设:S=y t=x
y'+ycosx=sin2x/2
y'+ycosx=sinxcosx
dy=sinxcosxdx-ycosxdx
dy= -sinxdsinx-ydsinx
sinx=t
dy=-tdt-ydt
dy/dt=-t-y
设u=t+y
dy=du-dt
du/dt-1=-u
du/dt=1-u
du/(u-1)=-dt
ln(u-1)=C0-t
(u-1)=e^(C0-t)
u=e^(C0-t)+1
y=u-t=e^(C0-t)-t+1
通解y=e^(C0-sinx)-sinx+1