y"+6y'+10y=(e^-3*x)/cosX*cosX
y"+6y'+10y=e^(-3x)/(cosx)^2 ,-π/2
A'(x)=[∫{0->x}du/cosu]/(cosx)^2
==>
A(x)=∫{0->x}dt/(cost)^2 [∫{0->t}du/cosu]= (交换积分次序)
=∫{0->x}du/cosu [∫{u->x}dt/(cost)^2]=
=∫{0->x}du/cosu [tgx-tgu]=
=tgx∫{0->x}du/cosu-∫{0->x}tgudu/cosu =
=tgxln(1/cosx+tgx)-1/cosx+1
4。 得(1)的1个特解为:
y=e^(-3x)[sinxln(1/cosx+tgx)-...全部
y"+6y'+10y=e^(-3x)/(cosx)^2 ,-π/2
A'(x)=[∫{0->x}du/cosu]/(cosx)^2
==>
A(x)=∫{0->x}dt/(cost)^2 [∫{0->t}du/cosu]= (交换积分次序)
=∫{0->x}du/cosu [∫{u->x}dt/(cost)^2]=
=∫{0->x}du/cosu [tgx-tgu]=
=tgx∫{0->x}du/cosu-∫{0->x}tgudu/cosu =
=tgxln(1/cosx+tgx)-1/cosx+1
4。
得(1)的1个特解为:
y=e^(-3x)[sinxln(1/cosx+tgx)-1+cosx]
所以(1)的通解为:
y=e^(-3x)[acosx+bsinx+sinxln(1/cosx+tgx)-1]。
。收起