Understanding Engineering Mathematics

1. Take the Laplace transform of the equation, inserting the initial values.


2. Solve for the transform of the solution, obtaining

y(s) ̃ =f(s)

3. Invert this transform to obtain the solution

y(t)=L−^1 [f(s)]

Note an important feature of the Laplace transform method concerning initial conditions.
In most elementary methods for solving differential equations we first find the general
solution, involving the appropriate number of arbitrary constants. These constants are then
found by applying the initial or boundary values, producing a number of equations which
are solved for the constants. In the Laplace (and other) transform method however the initial
conditions are automatically included, and there is no need to find the general solution.
The extension of this method to second and higher order initial value problems is
conceptually simple and merely requires more complicated manipulation and inversion.

Problem 17.9

Solve the initial value problem

y′′Yy= 0 y. 0 /=1, y′. 0 /= 0

by Laplace transform.

Taking the Laplace transform through the equation and applying the initial conditions gives

L[y′′]+L[y]=(s^2 + 1 )y ̃−s= 0

Hence

y(s) ̃ =

s

s^2 + 1

so

y(t)=L−^1

[

s

s^2 + 1

]

=cost

Exercises on 17.6

1. Solve the initial value problems

(i) y′+ 3 y= 2 y( 0 )= 4 (ii) y′−y=ty( 0 )= 0

Check by using alternative solutions, or substituting back in the equations.

2. Solve the initial value problem

y′′+ 3 y′+ 2 y= 20 e−^3 t y( 0 )=y′( 0 )= 0

Verify that your solution satisfies the equation and the initial conditions.