Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz5 => Topic started by: maoyafei on October 31, 2019, 01:08:49 PM

Question: y''+9y=9sec^2(3t) 0<t<pi/6
y''+9y=0
r^2+9=0
r1=3i
r2=3i
yc(t)=C1y1(t)+C2y2(t)
=C1cos3t+C2sin3t
W=cos3t*cos3tsin3t*(sin3t)=3
u1=integral of [(sin3t*9sec^2(3t))/3]dt
=integral of [3sin3t*(1/cos^2(3t)]dt
=3 integral of (sec3t*tan3t)dt
=  sec3t
=> u1=sec3t
u2=integral of [(cos3t*9sec^2(3t))/3]dt
=integral of [3 cost3t*(1/cos^2(3t)]dt
= ln/sec3t+tant3t/
=>u2 = ln/sec3t+tan3t/
yp(t)=u1y1+u2y2
=cos3t(sec3t)+sin3t*(ln/sec3t+tant3t/)
=sin3t(ln/sec3t+tant3t/)1
y(t)=yc(t)+yp(t)
=C1cos3t+C2sin3t+sin3t(ln/sec3t+tan3t/)1