b 2428 y y ea y 0 31 y 0 30 c 2430 y y sin ar y 0 0 y0 2 Sol

b) (24.28) y\" -y ea; y (0) 31, y (0) 30 c) (24.30) y\" y sin ar; y (0) 0, y\'(0) 2

Solution

b ) Given that

y\'\' - y = e^x , y(0) = 1 , y\'(0) = 0

Let y(s) = L [ y(x) ] (s)

y(x) = L^-1 ( y(s))

y\'\' - y = e^x

Take laplace transform on both sides

L [   y\'\' - y ] = L [ e^x ]

   L( y\'\' ) - L(y) = L ( e^x )

[ s²L(y) - sy(0) - y\'(0) ] -  L(y) =  L ( e^x )

[ s² y(s) - sy(0) - y\'(0) ] - y(s) = 1 / s -1 [ since, L(e^ax) = 1/(s - a) ]

y(s) [ s² - 1 ] - s(1) - 0 = 1 / s -1

  y(s) [ s² - 1 ] - s = 1 / s -1  

  y(s) [ s² - 1 ] = (  1 / s -1 ) + s

   y(s) [ s² - 1 ] = (s² -1) / (s-1) = (s² -1²) / (s-1)

= ( s+1) (s-1) / (s-1) [ since, (a² - b²) = ( a + b ) ( a - b) ]

= (s+1)

y(s) =  (s+1) / (s²-1)

= (s+1) / (s+1) (s-1)

= 1 / s-1

Hence,

y(x) = L^-1 ( y(s))

=  L^-1 ( 1 / s -1 )

y(x) = e^x [since,L^-1( 1/s-a) = e^ax ) ]

Therefore,

The solution is ,  y(x) = e^x

c ) Given that

y\'\' + y = sinx , y(0) = 0 ,y\'(0) = 2

Let y(s) = L [ y(x) ] (s)

y(x) = L^-1 ( y(s))

    y\'\' + y = sinx

Take laplace transform on both sides

L [   y\'\' + y ] = L [ sin x]

     L( y\'\' ) + L(y) = L ( sin x )

[ s²L(y) - sy(0) - y\'(0) ] + L(y) =  L ( sin x)

   [ s² y(s) - sy(0) - y\'(0) ] + y(s) = 1 /( s² + 1²) [ since, L(sin ax) = a /(s² + a²) ]

   [ s² y(s) - s(0) - 2 ] + y(s) = 1 /( s² + 1²)

  s² y(s) - 0 - 2 + y(s) = 1 /( s² + 1²)

  s² y(s) - 2 + y(s) = 1 /( s² + 1²)

y(s) [ s² - 1 ] -2 = 1 /( s² + 1²)

   y(s) [ s² - 1 ] = 1 /( s² + 1²) + 2

  y(s) [ s² - 1 ] = (3+2s²) / (s² + 1)

y(s) =  (3+2s²) / (s² + 1)(s² - 1)

Take partial fractions of y(s) ,

  y(s) =  (3+2s²) / (s² + 1)(s² - 1) = (3 + 2s²) / (s² + 1)(s + 1)(s - 1)

  (3+2s²) / (s² + 1)(s² - 1 ) = (As + B)/(s² + 1) + C/(s + 1) + D/(s - 1)  

On solving ,

A = -1/2 ,B = 0 , C = -5/4 ,D = 5/4

Hence,

y(s) =  (As + B)/(s² + 1) + C/(s + 1) + D/(s - 1)  

=  (-1/2s + 0)/(s² + 1) + (-5/4) / (s + 1) + (5/4)/(s - 1)

=    (-1/2s)/(s² + 1) + (-5/4) / (s + 1) + (5/4)/(s - 1)

Hence,

y(x) =  L^-1 ( y(s))

=   L^-1[ (-1/2s)/(s² + 1) + (-5/4) / (s + 1) + (5/4)/(s - 1) ]

= L^-1( (-1/2s)/(s² + 1) ) + L^-1(  (-5/4) / (s + 1) ) + L^-1( (5/4)/(s - 1) ) [ since,L^-1(s/s²+a²) =cosat ,

y(x) = (-1/2) cosx + (-5/4)e^-x + (5/4)e^x L^-1(1/s-a) = e^x ]

Therefore,

The solution is ,   (-1/2) cosx + (-5/4)e^-x + (5/4)e^x