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Question Number 99242 by abdomathmax last updated on 19/Jun/20
calculate L(e^(−ax) ch(3x)) with a>0
calculateL(e−axch(3x))witha>0
Commented by PRITHWISH SEN 2 last updated on 20/Jun/20
∫_0 ^∞ e^(−(s+a)x) ch(3x)dx=((s+a)/((s+a)^2 −9))
∫0∞e−(s+a)xch(3x)dx=s+a(s+a)2−9
Answered by mathmax by abdo last updated on 20/Jun/20
L(e^(−ax) ch(3x)) =∫_0 ^∞ e^(−at) ch(3t)e^(−xt) dt =(1/2)∫_0 ^∞ e^(−at) (e^(3t) +e^(−3t) )e^(−xt) dt =(1/2)∫_0 ^∞ (e^((−a+3−x)t) +e^((−a−3−x)t) )dt =(1/2)[(1/(3−a−x)) e^((3−a−x)t) +(1/(−3−a−x))e^((−3−a−x)t) ]_0 ^(+∞) =(1/2)(−(1/(3−a−x)) +(1/(3+a+x))) =(1/2)((1/(a+x+3)) +(1/(a+x−3))) =(1/2){((a+x−3+a+x+3)/((a+x)^2 −9))} =((a+x)/((a+x)^2 −9)) ⇒ L(e^(−ax) ch(3x)) =((x+a)/((x+a)^2 −9))
L(e−axch(3x))=∫0∞e−atch(3t)e−xtdt=12∫0∞e−at(e3t+e−3t)e−xtdt=12∫0∞(e(−a+3−x)t+e(−a−3−x)t)dt=12[13−a−xe(3−a−x)t+1−3−a−xe(−3−a−x)t]0+∞=12(−13−a−x+13+a+x)=12(1a+x+3+1a+x−3)=12{a+x−3+a+x+3(a+x)2−9}=a+x(a+x)2−9⇒L(e−axch(3x))=x+a(x+a)2−9

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