Question

Elementary Laplace Transtorms Y(S) = {f} -L e-stf(t)dt fc = C-'{F(s)} F(s) = {f} f(t) =-'{F(s)) F(s) = {f} -CS 1. 1 1 12. uct) le S> 0 S> 0 . s S 2. eat 1 13. ucOf(t-c) e-csF(s) S> a S-a n! 3. t",n e Z 14. ectf(t) F( sc) S> 0 sh+1 4. tP, p>-1 (p+1) S> 0 SP+1 15. f(ct) F). c>0 16. SFt - 1)g(t)dt F(s)G(*) 5. sin at S> 0 16. cos at 17. 8(t -c) S> 0 a s²+ a²1 S s²+ a² a s2 - 42' S s² - a²' 7. sinh at "F(5) - (0) ----(0) 18. f(n) (t) s> lal 8. cosh at f(n)(s) 19. (-t)"f(t) S> jal 9. eat sin bt 20. f(t + 7) = f(t) b (s-a)2 + b2' S> a 1 1 - e-s7 -st f(t)dt S-a 10. eat cos bt 21. f"(t) s-F(s) - sf (0) - f'(0) (s - a)2 + b2-5 > a 11. theat, n e Z+ n! 22. f'(o) SF(s)-f(0) sa

Answer 1

solution 6) The given initial value Problem is gy+ y = 1, Y0=0, U'O)= 0 Taking Laplace transform both side of the above differential equation, we have LY"+y} = ({0} or {"}+{4} = - - or $2 r(s)- Sy(0) - YO+Y(S) - or, ($2+1) Y(3) = 5 (40)=0 y'(o)=0 o=0) Y(S) = colo 1 1 or, s(s?+1) 1 or Y(S) = S 3+1 S CScanned with CamScanner

Non, taking inverse Laplace transform both side of the above equation, we get y (+) 1 cs or, y(t) = 27 Oro y(t) = 1- Cost Hence the required required solution of the given differential equation is y(t) = 1 - Cost IC Scanned with CamScanner