Laplace Transformation and Differential equations Mongi BLEL King - PowerPoint PPT Presentation

Laplace Transformation and Differential equations Mongi BLEL King Saud University March 17, 2020 Mongi BLEL Laplace Transformation and Differential equations

Exercise Use the Laplace transform to solve the initial-value problem y ′ + y = e − x + e x + cos x + sin x , y (0) = 1 . Mongi BLEL Laplace Transformation and Differential equations

Solution 1 : We begin by taking the Laplace transform of both sides to achieve + L [ y ] = L ( e − x + e x + cos x + sin x ) . � y ′ � L We know that 1 1 1 s L ( e − x + e x + cos x + sin x ) = s + 1 + s − 1 + s 2 + 1 + s 2 + 1. Denote Y ( s ) = L [ y ] , then 1 ( s + 1) 2 + 1 1 1 1 s Y ( s ) = s + 1 + 2( s − 1 − s + 1) + ( s + 1)( s 2 + 1) 1 + ( s + 1)( s 2 + 1) 1 1 1 1 = 2( s + 1) + ( s + 1) 2 + 2( s − 1) + s 2 + 1 . The solution of the differential equation: y = 1 2 e − x + xe − x + 1 2 e x + sin x . Mongi BLEL Laplace Transformation and Differential equations

Exercise Use Laplace transforms to solve the initial-value problem y ′ − 2 y = 5 + cos x + e 2 x + e − x , y (0) = 4 . Mongi BLEL Laplace Transformation and Differential equations

Solution 2 : By taking the Laplace transform of both sides of the differential equation, we get: sY − 4 − 2 Y = 5 1 1 s s + s 2 + 1 + s − 2 + s + 1 . Then 4 5 s 1 Y ( s ) = s − 2 + s ( s − 2) + ( s − 2)( s 2 + 1) + ( s − 2) 2 1 + ( s + 1)( s − 2) s − 2 + 5 4 � s − 2 − 1 1 � + 2 s − 2 + 1 1 − 2 s + 1 = s 2 + 1 2 5 5 s ( s − 2) 2 + 1 1 s − 2 − 1 1 1 + 3 3 s + 1 30( s − 2) − 5 217 2 s − 2 s 2 + 1 + 1 s 1 = s 2 + 1 5 5 ( s − 2) 2 − 1 1 1 + 3 s + 1 Mongi BLEL Laplace Transformation and Differential equations

and y ( x ) = 217 30 e 2 x − 5 2 − 2 5 cos x + 1 5 sin x + xe 2 x − 1 3 e − x . Mongi BLEL Laplace Transformation and Differential equations

Exercise Use the Laplace transform to solve the initial-value problem y ′′ − 2 y ′ + 2 y = cos x , y (0) = 1 , y ′ (0) = − 1 Mongi BLEL Laplace Transformation and Differential equations

Solution 3 : Using the Laplace transform of both sides of the differential equation, we get: s s 2 Y ( s ) − s + 1 − 2( sY ( s ) − 1) + 2 Y ( s ) = s 2 + 1 . s − 3 s Then Y ( s ) = ( s − 1) 2 + 1 + ( s 2 + 1)(( s − 1) 2 + 1) . Solving for Y ( s ) , we find: s − 1 ( s − 1) 2 + 1 + 1 2 s 2 + 1 − 2 s 1 Y ( s ) = ( s − 1) 2 + 1 − s 2 + 1 5 5 − 1 ( s − 1) 2 + 1 + 3 s − 1 1 ( s − 1) 2 + 1 . 5 5 Then e x cos x − 2 e x sin x + 1 5 cos x − 2 5 sin x − 1 5 e x cos x + 3 5 e x sin x = y 4 5 e x cos x − 7 5 e x sin x + 1 5 cos x − 2 = 5 sin x Mongi BLEL Laplace Transformation and Differential equations

Exercise Use Laplace transforms to solve the initial-value problem y ′ + y = 5 H ( x − 1) + e x H ( x − 1) + H ( x − 1) cos x , y (0) = 2 . Mongi BLEL Laplace Transformation and Differential equations

Solution 4 : Using the Laplace transform of both sides of the differential equation, we get: ( s + 1) Y − 2 = L [5 H ( x − 1) + e x H ( x − 1) + H ( x − 1) cos x ] . s e − s , L [ e x H ( x − 1)] = e − ( s − 1) Since L [ H ( x − 1)] = 1 and s − 1 L [ H ( x − 1) cos x ] = e − s sin 1 + cos 1 . We have: s 2 + 1 e − ( s − 1) 5 e − s ( s − 1)( s + 1) + (sin 1 + cos 1) e − s 2 Y ( s ) = s + 1 + s ( s + 1) + ( s + 1)( s 2 + 1) s + 1 + 5 e − s − e − s e − ( s − 1) 2 s + 1 + 1 = 2 s − 1 s e − ( s − 1) e − s − 1 s + 1 + (sin 1 + cos 1) 2 2 s + 1 se − s e − s − (sin 1 + cos 1) s 2 + 1 + (sin 1 + cos 1) s 2 + 1 2 2 Mongi BLEL Laplace Transformation and Differential equations

We have L − 1 � � 4 = 4 e − x , s +1 � 5 � L − 1 s e − s = 5 H ( x − 1) and � 5 � L − 1 s + 1 e − s = 5 H ( x − 1) e − ( x − 1) . Then y ( x ) = 4 e − x + 5 H ( x − 1) − 5 H ( x − 1) e − ( x − 1) . Mongi BLEL Laplace Transformation and Differential equations

Exercise Use the Laplace transform to solve the initial-value problem y ′′ + 2 y ′ + 5 y = H ( x − 2) , y (0) = 1 , y ′ (0) = 0 Mongi BLEL Laplace Transformation and Differential equations

Solution 5 : Using the Laplace transform of both sides of the differential equation, we get: s 2 Y ( s ) − sy (0) − y ′ (0) + 2( sY ( s ) − y (0)) + 5 Y ( s ) = e − 2 s . s = s + 2 + e − 2 s s 2 + 2 s + 5 � � Then Y ( s ) . Solving for Y ( s ) , we s find: s + 2 1 s 2 + 2 s + 5 + e − 2 s Y ( s ) = s ( s 2 + 2 s + 5) . s + 2 s + 1 1 We have s 2 + 2 s + 5 = ( s + 1) 2 + 4 + ( s + 1) 2 + 4 and � s + 1 � = e − x cos(2 x ) and L − 1 ( s + 1) 2 + 4 � 2 � = e − x sin(2 x ). L − 1 ( s + 1) 2 + 4 Mongi BLEL Laplace Transformation and Differential equations

e − 2 s s ( s 2 + 2 s + 5) = 1 � 1 s − ( s + 1) + 1 � 5 e − 2 s = ( s + 1) 2 + 4 1 � 1 s + 1 1 � 5 e − 2 s s − ( s + 1) 2 + 4 + ( s + 1) 2 + 4 . � � 1 s + 2 �� L − 1 e − 2 s s − = H ( x − 2)[1 s 2 + 2 s + 5 � � cos 2( x − 2) + 1 − e − ( x − 2) 2 sin 2( x − 2) ] . The solution y ( x ) to the initial-value problem is e − x cos(2 x ) + e − x sin(2 x ) + 1 y ( x ) = 5 H ( x − 2)[1 2 − e − ( x − 2) cos 2( x − 2) + e − ( x − 2) sin 2( x − 2)] 2 Mongi BLEL Laplace Transformation and Differential equations

Exercise Solve the differential equation y ′ + 3 y = 13 sin(2 t ) , y (0) = 6 . Mongi BLEL Laplace Transformation and Differential equations

Solution 6 : We take the Laplace transform of each member of the differential equation: 26 L ( y ′ )+3 L ( y ) = 13 L (sin(2 t )) . Then ( s +3) Y ( s ) − 6 = 6+ s 2 + 4. 6 26 s + 3 + − 2 s + 6 6 Y ( s ) = s + 3 + ( s + 3)( s 2 + 4) = and s 2 + 4 1 1 s y = 6 L − 1 ( s + 3) − 2 L − 1 ( s 2 + 4) + 6 L − 1 ( s 2 + 4) = 6 e − 3 t − 2 cos(2 t ) + 3 sin(2 t ) . Mongi BLEL Laplace Transformation and Differential equations

Exercise Solve the differential equation ′′ − 3 y ′ + 2 y = e − 4 t , y (0) = 1 , y ′ (0) = 5 . y Mongi BLEL Laplace Transformation and Differential equations

Solution 7 : We take the Laplace transform of each member of the differential equation: ′′ ) − 3 L ( y ′ ) + 2 L ( y ) = L ( e − 4 t ) . Then L ( y s 2 + 6 s + 9 ( s − 1)( s − 2)( s + 4 and y = − 16 5 e t + 25 5 e 2 t + 1 30 e − 4 t . Y ( s ) = Mongi BLEL Laplace Transformation and Differential equations

Exercise Solve the differential equation ′′ − 6 y ′ + 9 y = t 2 e 3 t , y (0) = 2 , y ′ (0) = 17 . y Solution 8 : We take the Laplace transform of each member of the differential equation: Y ( s ) = 2 s + 5 2 ( s − 3) 2 + ( s − 3) 5 . y = 2 e 3 t + 11 te 3 t 1 12 t 4 e 3 t . Mongi BLEL Laplace Transformation and Differential equations

Exercise Solve the following differential equation: y ′ − 2 y = f ( x ) , withy (0) = 3 , f ( x ) = 3 cos x for x ≥ 1 and f ( x ) = 0 , for 0 ≤ x ≤ 1 . Mongi BLEL Laplace Transformation and Differential equations

Solution 9 : s 2 +1 e − s and L ( f ( x )) = − 3 s s 2 +1 e − s . Then sF ( s ) − 3 − 2 F ( s ) = − 3 s 1 � 3 s � s 2 + 1 e − s F ( s ) = 3 − = s − 2 s − 2 + 6 3 s 2 + 1 e − s − 3 s 1 s 2 + 1 e − s . 5 5 L − 1 � � 3 = 3 e 2 t , s − 2 L − 1 � s 2 +1 e − s � 6 s = 6 5 cos( t − 1) H ( t − 1), 5 L − 1 � s 2 +1 e − s � 3 1 = 3 5 sin( t − 1) H ( t − 1). 5 y ( t ) = 3 e 2 t + 6 5 cos( t − 1) H ( t − 1) − 3 5 sin( t − 1) H ( t − 1) . Mongi BLEL Laplace Transformation and Differential equations

Solve the initial value problem y ′′ + y ′ + y = sin( x ) , y (0) = 1 , y ′ (0) = − 1 . Mongi BLEL Laplace Transformation and Differential equations

L{ y ′′ ( x ) } = s 2 Y ( s ) − sy (0) − y ′ (0) L{ y ′ ( x ) } = sY ( s ) − y (0) = sY ( s ) − 1 , Taking Laplace transforms of the differential equation, we get 1 ( s 2 + s + 1) Y ( s ) − s = s 2 + 1 . Then s 1 Y ( s ) = s 2 + s + 1 + ( s 2 + s + 1)( s 2 + 1) . √ s + 1 1 3) 2 − 1 3 s s 2 2 = = √ √ s 2 + s + 1 2 ) 2 + 3 2 ) 2 + ( 1 2 ) 2 + ( 1 ( s + 1 ( s + 1 ( s + 1 3 4 2 2 Mongi BLEL Laplace Transformation and Differential equations

Finding the inverse Laplace transform. Since √ s + 1 1 3 s s 3) 2 − 1 2 2 s 2 + s + 1 = = √ √ √ 2 ) 2 + 3 2 ) 2 + ( 1 2 ) 2 + ( 1 ( s + 1 ( s + 1 ( s + 1 3 3) 4 2 2 we have √ √ � s � 2 x cos(1 3 x ) − 1 2 x sin(1 = e − 1 e − 1 L − 1 √ 3 x ) . s 2 + s + 1 2 2 3 Also, we have 1 s + 1 s ( s 2 + s + 1)( s 2 + 1) = s 2 + s + 1 − s 2 + 1 , 1 1 s 2 + s + 1 = , and 2 ) 2 + 3 ( s + 1 4 √ � 1 � 2 2 x sin(1 e − 1 L − 1 √ = 3 x ) . s 2 + s + 1 2 3 Mongi BLEL Laplace Transformation and Differential equations

Then � 1 � � � � 1 � s L − 1 = L − 1 + L − 1 −L ( s 2 + s + 1)( s 2 + 1) s 2 + s + 1 s 2 + s + 1 We obtain √ 2 x cos(1 y ( x ) = 2 e − 1 3 x ) − cos( x ) . 2 Mongi BLEL Laplace Transformation and Differential equations

Solve the system of linear differential equation: � dx = − 2 x + y , dt with the initial conditions x (0) = 1, y (0) = 2. dy = x − 2 y dt Mongi BLEL Laplace Transformation and Differential equations