The Laplace Transform and the IVP (Sect. 6.2).

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The Laplace Transform and the IVP (Sect..2). Solving differential equations using L ]. Homogeneous IVP. First, second, higher order equations. Non-homogeneous IVP. Recall: Partial fraction decompositions. Solving differential equations using L ]. Remark: The method works with: Constant coefficient equations. Homogeneous and non-homogeneous equations. First, second, higher order equations. Idea of the method: ] differential eq. L for y(t). () Algebraic eq. for Ly(t)]. (2) (2) Solve the algebraic eq. for Ly(t)]. (3) Transform back to obtain y(t). (Using the table.)

Solving differential equations using L ]. Idea of the method: ] differential eq. L for y(t). (2) Solve the algebraic eq. for Ly(t)]. () (3) Algebraic eq. for Ly(t)]. Transform back to obtain y(t). (2) (Using the table.) Recall: (a) La f (t) + b g(t)] = a Lf (t)] + b Lg(t)]; (b) L y (n)] = s n Ly] s (n ) y(0) s (n 2) y (0) y (n ) (0). The Laplace Transform and the IVP (Sect..2). Solving differential equations using L ]. Homogeneous IVP. First, second, higher order equations. Non-homogeneous IVP. Recall: Partial fraction decompositions.

Homogeneous IVP. y y 2y = 0, y(0) =, y (0) = 0. Solution: Compute the L ] of the differential equation, Ly y 2y] = L0] Ly y 2y] = 0. The L ] is a linear function, so Ly ] Ly ] 2 Ly] = 0. Derivatives are transformed into power functions, ] ] s 2 Ly] s y(0) y (0) s Ly] y(0) 2 Ly] = 0, We the obtain (s 2 s 2) Ly] = (s ) y(0) + y (0). Homogeneous IVP. y y 2y = 0, y(0) =, y (0) = 0. Solution: Recall: (s 2 s 2) Ly] = (s ) y(0) + y (0). Differential equation for y Introduce the initial condition, L ] Algebraic equation for Ly]. (s 2 s 2) Ly] = (s ). We can solve for the unknown Ly] as follows, Ly] = (s ) (s 2 s 2).

Homogeneous IVP. y y 2y = 0, y(0) =, y (0) = 0. Solution: Recall: Ly] = (s ) (s 2 s 2). The partial fraction method: Find the zeros of the denominator, { s 2 s 2 = 0 s ± = ] s+ = 2, ± + 8 2 s =, Therefore, we rewrite: Ly] = Find constants a and b such that (s ) (s 2)(s + ). (s ) (s 2)(s + ) = a s 2 + b s +. Homogeneous IVP. Solution: Recall: y y 2y = 0, y(0) =, y (0) = 0. A simple calculation shows (s ) (s 2)(s + ) = a s 2 + (s ) (s 2)(s + ) = a s 2 + b s + b s +. a(s + ) + b(s 2) = (s 2)(s + ) (s ) = s(a + b) + (a 2b) { a + b =, a 2b = Hence, a = 3 and b = 2 3. Then, Ly] = 3 (s 2) + 2 3 (s + ).

Homogeneous IVP. y y 2y = 0, y(0) =, y (0) = 0. Solution: Recall: Ly] = 3 (s 2) + 2 3. From the table: (s + ) Le at ] = s a s 2 = Le2t ], s + = Le t ]. So we arrive at the equation Ly] = 3 Le2t ] + 2 3 Le t ] = L e 3( 2t + 2e t)] We conclude that: y(t) = 3( e 2t + 2e t). Homogeneous IVP. y 4y + 4y = 0, y(0) =, y (0) =. Solution: Compute the L ] of the differential equation, The L ] is a linear function, Ly 4y + 4y] = L0] = 0. Ly ] 4 Ly ] + 4 Ly] = 0. Derivatives are transformed into power functions, ] ] s 2 Ly] s y(0) y (0) 4 s Ly] y(0) + 4 Ly] = 0, Therefore, (s 2 4s + 4) Ly] = (s 4) y(0) + y (0).

Homogeneous IVP. y 4y + 4y = 0, y(0) =, y (0) =. Solution: Recall: (s 2 4s + 4) Ly] = (s 4) y(0) + y (0). Introduce the initial conditions, (s 2 4s + 4) Ly] = s 3. Solve for Ly] as follows: Ly] = (s 3) (s 2 4s + 4). The partial fraction method: Find the roots of the denominator, s 2 4s +4 = 0 s ± = 2 We obtain: Ly] = (s 3) (s 2) 2. 4± ] s + = s = 2. Homogeneous IVP. y 4y + 4y = 0, y(0) =, y (0) =. Solution: Recall: Ly] = (s 3) (s 2) 2. This expression is already in the partial fraction decomposition. Idea: Rewrite the right-hand side in terms of function in the table. Ly] = (s 2) + 2 3 (s 2) 2 = From the Laplace transforms table: Le at ] = s a Lt n e at n! ] = (s a) (n+) (s 2) (s 2) 2 (s 2) 2 = s 2 (s 2) 2. s 2 = Le2t ], (s 2) 2 = Lte2t ].

Homogeneous IVP. y 4y + 4y = 0, y(0) =, y (0) =. Solution: Recall: Ly] = s 2 (s 2) 2 and s 2 = Le2t ], (s 2) 2 = Lte2t ]. So we arrive at the equation Ly] = Le 2t ] Lte 2t ] = L e 2t te 2t]. We conclude that y(t) = e 2t te 2t. The Laplace Transform and the IVP (Sect..2). Solving differential equations using L ]. Homogeneous IVP. First, second, higher order equations. Non-homogeneous IVP. Recall: Partial fraction decompositions.

First, second, higher order equations. Use the Laplace Transform to find the solution of y (4) 4y = 0, y(0) =, y (0) =, y (0) = 2, y (0) = 0. Solution: Compute the L ] of the equation, L y (4)] 4 Ly] = 0. s 4 Ly] s 3 y(0) s 2 y (0) s y (0) y (0) ] 4 Ly] = 0. s 4 Ly] s 3 + 2s ] 4 Ly] = 0 (s 4 4) Ly] = s 3 2s, We obtain, Ly] = s3 2s (s 4 4). First, second, higher order equations. Use the Laplace Transform to find the solution of y (4) 4y = 0, y(0) =, y (0) =, y (0) = 2, y (0) = 0. Solution: Recall: Ly] = Ly] = s3 2s (s 4 4). s(s 2 2) (s 2 2)(s 2 + 2) Ly] = s (s 2 + 2). The last expression is in the table of Laplace Transforms, Ly] = s ( s 2 + 2 ] 2) = L cos( 2 t) ]. We conclude that y(t) = cos( 2 t).

The Laplace Transform and the IVP (Sect..2). Solving differential equations using L ]. Homogeneous IVP. First, second, higher order equations. Non-homogeneous IVP. Recall: Partial fraction decompositions. Non-homogeneous IVP. y 4y + 4y = 3 sin(2t), y(0) =, y (0) =. Solution: Compute the Laplace transform of the equation, Ly 4y + 4y] = L3 sin(2t)]. The right-hand side above can be expressed as follows, L3 sin(2t)] = 3 Lsin(2t)] = 3 2 s 2 + 2 2 = Introduce this source term in the differential equation, Ly ] 4 Ly ] + 4 Ly] =

Non-homogeneous IVP. y 4y + 4y = 3 sin(2t), y(0) =, y (0) =. Solution: Recall: Ly ] 4 Ly ] + 4 Ly] = Derivatives are transformed into power functions, s 2 Ly] s y(0) y (0) Rewrite the above equation, ] 4 s Ly] y(0) ] + 4 Ly] = (s 2 4s + 4) Ly] = (s 4) y(0) + y (0) + Introduce the initial conditions, (s 2 4s + 4) Ly] = s 3 + Non-homogeneous IVP. y 4y + 4y = 3 sin(2t), y(0) =, y (0) =. Solution: Recall: (s 2 4s + 4) Ly] = s 3 + Therefore, Ly] = (s 3) (s 2 4s + 4) + (s 2 4 + 4)(s 2 + 4). From an above: s 2 4s + 4 = (s 2) 2, Ly] = s 2 (s 2) 2 + (s 2) 2 (s 2 + 4). From an above we know that Le 2t te 2t ] = s 2 (s 2) 2.

Non-homogeneous IVP. y 4y + 4y = 3 sin(2t), y(0) =, y (0) =. Solution: Recall: Ly] = Le 2t te 2t ] + (s 2) 2 (s 2 + 4). Use Partial fractions to simplify the last term above. Find constants a, b, c, d, such that (s 2) 2 (s 2 + 4) = as + b s 2 + 4 + c (s 2) + d (s 2) 2 (s 2) 2 (s 2 + 4) = (as + b)(s 2)2 + c(s 2)(s 2 + 4) + d(s 2 + 4) (s 2 + 4)(s 2) 2 = (as + b)(s 2) 2 + c(s 2)(s 2 + 4) + d(s 2 + 4). Non-homogeneous IVP. y 4y + 4y = 3 sin(2t), y(0) =, y (0) =. Solution: = (as + b)(s 2) 2 + c(s 2)(s 2 + 4) + d(s 2 + 4). = (as + b)(s 2 4s + 4) + c(s 3 + 4s 2s 2 8) + d(s 2 + 4) = a(s 3 4s 2 +4s)+b(s 2 4s +4)+c(s 3 +4s 2s 2 8)+d(s 2 +4). We obtain the system = (a + c)s 3 + ( 4a + b 2c + d)s 2 + (4a 4b + 4c)s + (4b 8c + 4d). a + c= 0, 4a + b 2c + d= 0, 4a 4b + 4c= 0, 4b 8c + 4d =.

Non-homogeneous IVP. y 4y + 4y = 3 sin(2t), y(0) =, y (0) =. Solution: The solution for this linear system is a = 3 8, b = 0, c = 3 8, d = 3 4. (s 2) 2 (s 2 + 4) = 3 8 s s 2 + 4 3 8 Use the table of Laplace Transforms (s 2) + 3 4 (s 2) 2. (s 2) 2 (s 2 + 4) = 3 8 Lcos(2t)] 3 8 Le2t ] + 3 4 Lte2t ]. (s 2) 2 (s 2 + 4) = L 3 8 cos(2t) 3 8 e2t + 3 4 te2t]. Non-homogeneous IVP. y 4y + 4y = 3 sin(2t), y(0) =, y (0) =. Solution: Summary: Ly] = Le 2t te 2t ] + (s 2) 2 (s 2 + 4), (s 2) 2 (s 2 + 4) = L 3 8 cos(2t) 3 8 e2t + 3 4 te2t]. Ly(t)] = L ( t) e 2t + 3 8 ( + 2t) e2t + 3 8 cos(2t) ]. We conclude that y(t) = ( t) e 2t + 3 8 (2t ) e2t + 3 8 cos(2t).

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