Question

angulo y11 - y=1 ; y(0)=0. y'(0)= 1/2

Answer 1

         $L\{y''-y\}=L\{1\}\\ \\ L\{y''\}-L\{y\}=\dfrac{1}{s}\\ \\ s^2L\{y\}-s\cdot y(0)-y'(0)-L\{y\}=\dfrac{1}{s}\\ \\ (s^2-1)L\{y\}-\dfrac{1}{2}=\dfrac{1}{s}\\ \\ L\{y\}=\dfrac{1}{s(s^2-1)}+\dfrac{1}{2(s^2-1)}\\ \\ \\ L\{y\}=\dfrac{1}{s(s+1)(s-1)}+\dfrac{1}{2}\cdot\dfrac{1}{(s+1)(s-1)}\\ \\ \\ L\{y\}=\dfrac{3}{4(s-1)}+\dfrac{1}{4(s + 1)}-\dfrac{1}{s}$
   
         $y=L^{-1}\left\{\dfrac{3}{4(s-1)}+\dfrac{1}{4(s + 1)}-\dfrac{1}{s}\right\}\\ \\ \\ y=\dfrac{3}{4}L^{-1}\left\{\dfrac{1}{s-1}\right\}+\dfrac{1}{4}L^{-1}\left\{\dfrac{1}{s+1}\right\}-L^{-1}\left\{\dfrac{1}{s}\right\}\\ \\ \\ \boxed{y=\dfrac{3}{4}e^{x}+\dfrac{1}{4}e^{-1}-1}$