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Applications of the Laplace transform in solving ordinary differential equations

1. Ordinary differential equations with constant coefficients. The Laplace transform is useful in solving linear ordinary differential equations with constant coefficients. Consider the second order linear differential equation

or

2) Y" + aY' + bY = G(t)

subject to the initial conditions

Y(0) = A, Y' (0) = B

where a, b, A, B are constants. We note that Y is shorthand for Y(t). Let the Laplace transform of Y(t) be y(s), or, more concisely, y. If we take the Laplace transform of 2) and employ the theorems for the Laplace transform of derivatives we obtain an algebraic equation in the variables y and s. If we then solve for y in terms of s and take the inverse transform, we obtain the desired solution Y.

Example 1. Solve the equation

3) Y" - 3Y' + 2Y = 4e^{2 t}

where Y(0) =-3, Y'(0) = 5

Solution. Y is shorthand for Y(t). Let the Laplace transform of Y(t) be y(s), or, more concisely, y.

Taking the Laplace transform of both sides of 3), we obtain

L[Y"} - 3 L[Y'] + 2 L[Y] = 4 L[e^{2 t}]

Using the theorems for the transforms of derivatives, we get

Substituting in the values of Y(0) = -3, Y'(0) = 5 we get

The solution is then

Example 2. Solve the equation

Y''' - 3Y" + 3Y' - Y = t^{2}e^{ t}

where Y(0) =1, Y'(0) = 0, Y"(0) = -2.

Solution. Y is shorthand for Y(t). Let the Laplace transform of Y(t) be y(s), or, more concisely, y.

Taking the Laplace transform, we obtain

L[Y'''] - 3 L[Y"] + 3 L[Y'] - L[Y] = L[t^{2}e^{ t}]

Using the theorems for the transforms of derivatives, we get

{s^{3}y - s^{2} Y(0) - s Y'(0) - Y"(0)} - 3{s^{2}y - s Y(0) - Y'(0)} + 3{sy - Y(0)} - y

Substituting in the values of Y(0) =1, Y'(0) = 0, Y"(0) = -2, we obtain

Now s^{3} - 3s^{2} + 3s - 1 = (s - 3)^{3}, so we get

Taking the inverse transform, we then get

Example 3. Find the general solution of the differential equation of Example 2:

Y''' - 3Y" + 3Y' - Y = t^{2}e^{ t}

Let Y(0) =A, Y'(0) = B, Y"(0) = C.

Solution. Y is shorthand for Y(t). Let the Laplace transform of Y(t) be y(s), or, more concisely, y.

Taking the Laplace transform, we obtain

L[Y'''] - 3 L[Y"] + 3 L[Y'] - L[Y] = L[t^{2}e^{ t}]

Using the theorems for the transforms of derivatives, we get

{s^{3}y - s^{2} Y(0) - s Y'(0) - Y"(0)} - 3{s^{2}y - s Y(0) - Y'(0)} + 3{sy - Y(0)} - y

Substituting in the values of Y(0) = A, Y'(0) = B, Y"(0) = C, we get

Now s^{3} - 3s^{2} + 3s - 1 = (s - 3)^{3}, so we have

Since A, B and C are arbitrary, the polynomial in the numerator of the first term on the right is also arbitrary. Thus

Inverting we obtain the required general solution

where the c_{k}’s are arbitrary constants.

Example 4. Solve the equation

Y" + a^{2}Y = F(t)

where Y(0) = 1, Y'(0) = -2.

Solution. Y is shorthand for Y(t). Let the Laplace transform of Y(t) be y(s), or, more concisely, y.

Taking the Laplace transform, we obtain

L[Y"] + a^{2} L[Y] = L[F(t)]

Using the theorems for the transforms of derivatives and denoting the Laplace transform of F(t) by f(s), we obtain

{s^{2}y - sY(0) - Y'(0)} + a^{2}y = f(s)

Substituting in the values of Y(0) =1, Y'(0) = -2, we obtain

s^{2}y - s + 2 + a^{2}y = f(s)

So

Taking the inverse

Using the convolution theorem

Example 5. Find the general solution of

Y" + a^{2}Y = F(t)

Solution. Let Y(0) = c_{1}, Y'(0) = c_{2}.

Taking the Laplace transform, we obtain

L[Y"] + a^{2} L[Y] = L[F(t)]

Using the theorems for the transforms of derivatives and denoting the Laplace transform of F(t) by f(s), we obtain

{s^{2}y - sY(0) - Y'(0)} + a^{2}y = f(s)

Substituting in the values of Y(0) = c_{1}, Y'(0) = c_{2}, we obtain

s^{2}y - sc_{1} - c_{2} - a^{2}y = f(s)

Inverting we get

where A and B are arbitrary constants.

2. Ordinary differential equations with variable coefficients. The Laplace transform can be used in solving some ordinary differential equations with variable coefficients. One particular equation where the method is useful is one in which the terms have the form

t^{m} Y^{(n)}(t)

the Laplace transform of which is

Example 6. Solve the equation

tY" + Y' + 4tY = 0

where Y(0) = 3, Y'(0) = 0.

Solution. Y is shorthand for Y(t). Let the Laplace transform of Y(t) be y(s), or, more concisely, y.

Taking the Laplace transform, we obtain

L[tY"] + L [Y'] +^{ } L[4tY] = 0

Using the theorems for the transforms of derivatives, we obtain

Substituting in the values of Y(0) = 3, Y'(0) = 0, we obtain

Integrating

or

Inverting

Y = cJ_{0}(2t)

To determine c we note that Y(0) = cJ_{0}(0) = c = 3

Thus

Y = 3J_{0}(2t)

3. Simultaneous ordinary differential equations. The Laplace transform can be used to solve two or more simultaneous ordinary differential equations. The procedure is essentially the same as the one used above.

Example 7. Solve the system

where X(0) = 8, Y(0) = 3.

Solution. X is shorthand for X(t) and Y is shorthand for Y(t). Let the Laplace transform of X(t) be x(s), or, more concisely, x. Let the Laplace transform of Y(t) be y(s), or, more concisely, y.

Taking the Laplace transform, we get

sx - X(0) = 2x - 3y

sy - Y(0) = y - 2x

Substituting in X(0) = 8, Y(0) = 3, we get

sx - 8 = 2x - 3y

sy - 3 = y - 2x

or

(s - 2)x + 3y = 8

2x + (s -1)y = 3

Solving this system simultaneously,

Taking inverses,

X = L^{-1}[x] = 5e^{- t} + 3e^{4t}

Y = L^{-1}[y] = 5e^{- t} + 2e^{4t}

References

Murray R. Spiegel. Laplace Transforms. (Schaum)

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