All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 145773 by help last updated on 07/Jul/21
Answered by mathmax by abdo last updated on 08/Jul/21
→4r2+24r+37=0Δ′=122−4×37=144−148=−4⇒r1=−12+2i4=−3+12iandr2=−12−2i4=−3−12iy(x)=aer1x+ber2x=ae(−3+i2)x+be(−3+i2)x=e−3x(αcos(x2)+βsin(x2))y(π)=1⇒e−3π(α0+β)=1⇒β=e3πy′(x)=−3e−3x(αcos(x2)+βsin(x2))+e−3x(−α2sin(x2)+β2cos(x2))y′(π)=0⇒−3e−3π(β)+e−3π(−α2)=0⇒−3−e−3π2α=0⇒e−3π2α=3⇒α=6e3π⇒y(x)=e−3x(6e3πcos(x2)+e3πsin(x2))=e3(π−x){cos(x2)+sin(x2)}
Terms of Service
Privacy Policy
Contact: info@tinkutara.com