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Question Number 96773 by abdomathmax last updated on 04/Jun/20
solve y^(′′) −y =((sinx)/x)
solvey″−y=sinxx
Answered by abdomathmax last updated on 05/Jun/20
let solve by laplace (e)⇒L(y^(′′) )−L(y) =L(((sinx)/x)) ⇒ x^2 L(y)−xy(0)−y^′ (0) −L(y) =L(((sinx)/x)) ⇒ (x^2 −1)L(y) =xy(0)+y^′ (0)+L(((sinx)/x)) we have L(((sinx)/x)) =∫_0 ^∞ ((sint)/t) e^(−xt) dt =f(x) f^′ (x) =−∫_0 ^∞ sint e^(−xt) dt =−Im(∫_0 ^∞ e^(it−xt) dt) ∫_0 ^∞ e^((−x+i)t) dt =[(1/(−x+i)) e^((−x+i)t) ]_0 ^∞ =−(1/(−x+i)) =(1/(x−i)) =((x+i)/(x^2 +1)) ⇒f^′ (x) =−(1/(1+x^2 )) ⇒ f(x)=−arrctanx +c f(0) =(π/2) =0+c ⇒f(x) =(π/2) −arcrtanx e ⇒(x^2 −1)L(y) =xy(0)+y^′ (0)+(π/2)−arctanx ⇒L(y) =y(0)(x/(x^2 −1)) +((y^′ (0)+(π/2))/(x^2 −1))−((arctanx)/(x^2 −1)) y(x) =y(0)L^(−1) ((x/(x^2 −1)))+(y^′ (0)+(π/2))L^(−1) ((1/(x^2 −1))) −L^(−1) (((arctanx)/(x^2 −1))) (x/(x^2 −1)) =(x/((x−1)(x+1))) =(a/(x−1)) +(b/(x+1)) ⇒ a =(1/2) and b =(1/2) ⇒L^(−1) ((x/(x^2 −1))) =(1/2)e^x +(1/2)e^(−x) =ch(x) (1/(x^2 −1)) =(1/2)((1/(x−1))−(1/(x+1))) ⇒ L^(−1) ((1/(x^2 −1))) =(e^x /2)−(e^(−x) /2) =sh(x) rest to find L^(−1) (((arctanx)/(x^2 −1)))=δ(x) ⇒ y(x) =y(o)ch(x)+(y^′ (0)+(π/2))shx +δ(x)
letsolvebylaplace(e)⇒L(y″)−L(y)=L(sinxx)⇒x2L(y)−xy(0)−y′(0)−L(y)=L(sinxx)⇒(x2−1)L(y)=xy(0)+y′(0)+L(sinxx)wehaveL(sinxx)=∫0∞sintte−xtdt=f(x)f′(x)=−∫0∞sinte−xtdt=−Im(∫0∞eit−xtdt)∫0∞e(−x+i)tdt=[1−x+ie(−x+i)t]0∞=−1−x+i=1x−i=x+ix2+1⇒f′(x)=−11+x2⇒f(x)=−arrctanx+cf(0)=π2=0+c⇒f(x)=π2−arcrtanxe⇒(x2−1)L(y)=xy(0)+y′(0)+π2−arctanx⇒L(y)=y(0)xx2−1+y′(0)+π2x2−1−arctanxx2−1y(x)=y(0)L−1(xx2−1)+(y′(0)+π2)L−1(1x2−1)−L−1(arctanxx2−1)xx2−1=x(x−1)(x+1)=ax−1+bx+1⇒a=12andb=12⇒L−1(xx2−1)=12ex+12e−x=ch(x)1x2−1=12(1x−1−1x+1)⇒L−1(1x2−1)=ex2−e−x2=sh(x)resttofindL−1(arctanxx2−1)=δ(x)⇒y(x)=y(o)ch(x)+(y′(0)+π2)shx+δ(x)

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