MAT 341 Note the seventh lecture note PDF

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This is the lecture note that was given to us during one of the classes for mathematical methods in mat 351...

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MAT 341: MATHEMATICAL METHODS1. Applications of Laplace Transformations to Solutions of Ordinary Differential Equations (ODE). 17TH AND 18TH MAY, 2021.

1

Second Order o.d.e

Laplace transformations can be applied in the solution of ordinary differential equations with given initial conditions. Example 1.1 Solve the Initial Value Problem (IVP) y ′′ − 4y = 0. y ′ (0) = 2.

y(0) = 1, Solution

Recall that the Laplace transformation for second derivative is given by: L(f ′′ )(s) = s2 L(f )(s) − sf (0) − f ′ (0). Apply Laplace tranformation on the differential equation and use the given conditions, we obtain, L(y ′′ − 4y) = L(0) L(y ′′ ) − 4L(y) = 0 s2 L(y) − sy(0) − y ′ (0) − 4L(y) = 0 Now let

L(y) = y¯ s2 y¯ − s.1 − 2 − 4¯ y=0 (s2 − 4)¯ y = s+2 s+2 y¯ = 2 s −4  1  therefore y = L−1 s−2 ⇒ y(t) = e2t

Let’s consider the case of non-homogeneous ode, 1

Example 1.2 Solve the Initial Value Problem (IVP) y ′′ − 5y ′ + 4y = e2t . y′ (0) = 0.

y(0) = 1, Solution

Apply Laplace tranformation on the differential equation and use the given conditions, we obtain, L(y ′′ − 5y ′ + 4y) = L(e2t )

1 s−2 1 s2 L(y) − sy(0) − y ′ (0) − 5[sL(y ) − y(0)] + 4L(y ) = s−2 L(y ′′) − 5L(y ′ ) + 4L(y) =

Now let

L(y) = y¯

1 s−2 1 2 y¯(s − 5s + 4) − s + 5 = s−2 s2 − 7s + 11 y¯ = (s − 2)(s2 − 5s + 4) s2 − 7s + 11 y¯ = (s − 2)(s − 1)(s − 4)

s2 y¯ − s.1 − 0 − 5[s¯ y − 1] + 4¯ y=

Therefore

1 1  5 1  1 1  + − 2 s−2 3 s−1 6 s−4 1 2t 5 t 1 4t ⇒y=− e + e − e . 2 3 6 Exercise 1.1 Solve the following IVP by Laplace Transformation: y¯ = −

1.

y ′′ + 2y ′ + 5y = 0

y(0) = 2,

y ′ (0) = 4.

2.

y ′′ − 4y ′ + 4y = t2

y(0) = 0,

y ′ (0) = 1.

3.

y ′′ + 2y ′ + 2y = 0

y(0) = 1,

y ′ (0) = 1.

2

Simultaneous ODE

Some steps to note in solving simultaneous ode: 1. Take the Laplace transformation of both equations; 2. Insert the initial conditions and 3. Solve for x ¯ and y¯ by normal algebraic method. 2

Example 2.1 Solve the pair of simultaneous equations: y˙ − x = et x˙ + y = e−t , given that at t = 0;

x = 0, y = 0.

Solution Take the Laplace transformations of the equations, we obtain 1 s−1   1 s¯ x − x0 + y¯ = s+1 

 s¯ y − y0 − x¯ =

Solve the equations for x¯ and y¯, then take the inverse Laplace transformations; we obtain 1 1 x = − et − e−t + cos t + sin t 2 2 1 t 1 −t y = e + e − cos t + sin t. 2 2 Example 2.2 Solve the pair of simultaneous equations: x¨ + 2x − y = 0 y¨ + 2y − x = 0, given that at t = 0; x0 = 4, y0 = 2; x1 = 0, y1 = 0. Solution Take the Laplace transformations of the equations, we obtain  x − y¯ = 0 s2x¯ − sx0 − x1 + 2¯   2 y − x¯ = 0 s y¯ − sy0 − y1 + 2¯



Using initial conditions; we obtain s2x¯ − 4s − +2¯ x − y¯ = 0 s2 y¯ − 2s + 2¯ y − x¯ = 0 Simplify and take the inverse Laplace transformation, we obtain √ x = 3 cos t + cos( 3t) √ y = 3 cos t − cos( 3t) Exercise 2.1 Solving the following pairs of ode by Laplace transformation 1.

x˙ + y˙ + x + 2y = e−3t x˙ + 3x + 5y = 5e−2t 3

given that at t = 0; x = 4 and y = −1. 2.

2x˙ + 2x + 3y˙ + 6y = 56et − 3e−t x˙ − 2x − y¯ − 3y = −21et − 7e−t

given that at t = 0; x = 8 and y = 3. 3.

5¨ x + 12¨ y + 6x = 0 5¨ x + 16¨ y + 6y = 0

given that at t = 0; x =

7 4

, y = 1, x1 = 0, y1 = 0.

4...