f t te e = possesses a Laplace transform. Exercises for Module-III (Transform Calculus)

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1 Exercises for Module-III (Transform Calculus) ) Discuss he piecewise coninuiy of he following funcions: =,, +, > c) e,, = d) sin,, = ) Show ha he funcion ( ) sin ( ) f e e = possesses a Laplace ransform. 3) Find Laplace ransform of he following funcions: e cos, >, < ea, = > d) f( ) = H(, > c) f( ) sin, 4) Find he inverse Laplace ransform of he following funcions: Fs () = s e c) Fs () = s s + 3 ( s + 6s+ 3) d) Fs () = Fs () = 4s + 5 ( s ) ( s+ ) s + 3 ( s+ ) ( s + ) 5) Solve he following iniial value problems for > using Laplace ransform mehod: + y = ; y() = y'() = d dy + = H ( ) ; y() =, y'() = d d + y = f( ); y() = y'() = ; where f( ) = d ( ) c) cos,, >

2 6) Solve he following boundary value problems using Laplace ransform mehods: + y = sin, y() =, y, = d + 9 λ =, y() =, y' = d 3 7) Solve he following differenial equaions using Laplace ransform mehod: dy = ; y() = d d dy + ( + ) + y = e ; y() = d d 8) Solve he following inegral equaions using Laplace ransform mehod: y ( ) = sin + yu ( )cos( u)du dy() + 3 ( ) + ( )d = ; = d y yu u y( ) 9) Wih he applicaion of Laplace ransform, solve he Hea equaion u u = ; x>, > subjec o he condiions ux (, + ) =, x> u(, ) = f( ), > lim ux (, ) = x ) Using Laplace ransform solve he wave equaion wih he iniial and boundary condiions y y = ; < x<, > yx (, + ) = sin x, < x<, y ( x, + ) =, < x< y(, ) =, >, y(, ) =, >

3 x ) Expand f( x) = e, < x< in a Fourier series. ) Find he half range sine series of he funcion f( x) = x, < x< 3) Deermine he Fourier inegral represenaion of he funcion sin x, < x< f( x) =., x< and x> 4) Le he funcion f( x ) is given as x, < x< f( x) =., < x< Deermine he Fourier sine inegral of f( x ) and find he value of ( sin ) sin ( ) d. 5) Find he Fourier sine ransform of f ( ) = e,. 6) Deermine he Fourier cosine ransform for f( ) = e cos, and show ha ( + ) cos e cos = d,. 7) Find Fourier ransform of ff() = exp( aa ). 8) Find ff() if Fs ( ) =. + 9) Solve he hea equaion u u = k, < x<subjec o he condiions ux (,) = e x ; ux (, ) =, >, u and ux boh end o zero as x. ) Solve he following Laplace equaion condiions ux and u as x + y and u u + = ; y > subjec o he y ux (,) =, x, x >.

4 Answers/Hins: ) ) Funcion is no piecewise coninuous since lim f( ) do no exiss. ± Funcion is coninuous everywhere. c) Funcion has a jump disconinuiy a = piecewise coninuous. d) Funcion is coninuous everywhere. Definiion of Laplace ransform and inegraion by pars give Noe ha { cos( )} s { ( )} ( ) ( ) sin = d = cos() cos and hence he funcion is { } L f e e e sl e L e exiss because cos( e ) Hence L{ } exiss. 3) L{ e cos } ( ) ( s ) ( s ) 3 s + + s + s+ = = a ( ) c) { } { } ( s+ )( s + s+ 5) ( ) ( ) a ( s a = e H = e e H = e s a s s + e L f( ) = e sin d = s e e s + as as e e s s s d) L{ f () } = e + a = [ + as] as s ( )( ) is coninuous and is of exponenial order. 4) s+ 3 s+ 3 3 s 3 L L e L e sin = = = ( s + 6s+ 3) ( s + 3) + 4 ( s + 4) 4 e + 3e e 3 3 ( )sinh{ } H e + e cos + sin sin c) ( ) d)

5 5) y ( ) = cos + y = + e H( )( + e ) c) y = sin + H( )( ) sin ( ) 6) y = cos + cos + sin 4 y = + cos3+ sin ) y = + c y = e 8) y() = e 5 y () = e + e 9) x x ux (, ) = exp f( u)du ( ) 4( u) 3 u ) yx (, ) = sinxcos ) e f( x) cos x cos x... sin x sin x

6 ) 4 4 f( x) sin x sin x+ sin 3x sin 4 x ] 3) sin sin x d 4) sin f( x) sin x d and he value is. 4 5) ( + ) 6) + 4 (4 + ) 7) ia ( a + ) 8) f() = exp ( ) 9) k ux (, ) = e cos xd + ) y x x+ uxy (, ) = an + an x y y