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# 8. Using Inverse Laplace Transforms to Solve Differential Equations

## Laplace Transform of Derivatives

We use the following notation:

Subsidiary Equation

Application

(a) If we have the function g(t), then G(s) = G = Lap{g(t)}.

(b) g(0) is the value of the function g(t) at t = 0.

(c) g'(0), g’’(0),... are the values of the derivatives of the function at t = 0.

If g(t) is continuous and g'(0), g’’(0),... are finite, then we have the following.

### First Derivative

Lap{g"'"(t)}= Lap{(dg)/(dt)} =sG-g(0)

### Second Derivative

Lap{g"''"(t)}=s^2G-s\ g(0) - g"'"(0)

We saw many of these expressions in the Table of Laplace Transforms.

NOTATION NOTE:
If instead of g(t) we have a function y of x, then Equation (2) would simply become:

Lap{y’’(x)} = s^2Y − s\ y (0) − y’(0)

Likewise, if we have an expression for current i and it is a function of t, then the equation would become:

Lap{i’’(t)} = s^2I − s\ i(0) − i\ ’(0)

### For the n-th derivative

Lap{(d^ng)/(dt^n)} =s^nG-s^(n-1)g(0)-s^(n-2)g ’ (0)- ...-g^(n-1)(0)

NOTATION NOTE: If we have y and it is a function of t, then the notation would become:

Lap{(d^ny)/(dt^n)} =s^nY-s^(n-1)y(0)-s^(n-2)y ’ (0)- ...-y^(n-1)(0)

## Subsidiary Equation

The subsidiary equation is the equation in terms of s, G and the coefficients g'(0), g’’(0),... etc., obtained by taking the transforms of all the terms in a linear differential equation.

The subsidiary equation is expressed in the form G = G(s).

## Examples

Write down the subsidiary equations for the following differential equations and hence solve them.

### Example 1

(dy)/(dt)+y=sin\ 3t, given that y = 0 when t = 0.

Taking Laplace transform of both sides gives:

(sY-y(0))+Y=3/(s^2+9)

sY+Y=3/(s^2+9) (since y(0) = 0)

(s+1)Y=3/(s^2+9)

Solving for Y and finding the partial fraction decomposition gives:

Y=3/((s+1)(s^2+9)) =A/(s+1)+(Bs+C)/(s^2+9)

3=A(s^2+9)+(s+1)(Bs+C)

Substituting convenient values of s gives us:

s=-1 gives 3=10A, which gives A=3/10.

s=0 gives 3=9A+C, which gives C=3/10.

s=1 gives 3=10A+2B+2C, which gives us B=-3/10.

So

Y=3/((s+1)(s^2+9))

=3/10(1/(s+1)+(-s+1)/(s^2+9))

=3/10(1/(s+1)-s/(s^2+9)+1/(s^2+9))

Finding the inverse Laplace tranform gives us the solution for y as a function of t:

y=3/10e^(-t)-3/10cos\ 3t+1/10sin\ 3t

#### Solution Graph for Example 1

This is the graph of the solution we obtained in the example above.

### Example 2

Solve (d^2y)/(dt^2)+2(dy)/(dt)+5y=0, given that y = 1, and (dy)/(dt)=0, when t = 0.

Taking Laplace transform of both sides and appying initial conditions of y(0) = 1 and y"'"(0) = 0 gives:

{s^2Y-sy(0)-y"'"(0)}+ 2{sY- y(0)}+5Y= 0

(s^2Y-s)+2(sY-1)+5Y=0

(s^2+2s+5)Y=s+2

Solving for Y and completing the square on the denominator gives:

Y=(s+2)/(s^2+2s+5)

=(s+2)/((s^2+2s+1)+4)

=(s+2)/((s+1)^2+4)

=(s+1)/((s+1)^2+4)+1/2 2/((s+1)^2+4)

Now, finding the inverse Laplace Transform gives us the solution for y as a function of t:

y=e^(-t)cos\ 2t+1/2e^(-t) sin\ 2t

#### Solution Graph for Example 2

Here is the graph of what we just found:

### Example 3

(d^2y)/(dt^2)-2(dy)/(dt)+y=e^t, given that y = -2, and (dy)/(dt)=-3 when t = 0.

Taking Laplace transform of both sides:

{s^2Y-sy(0)-y"'"(0)}- 2{sY-y(0)}+Y =1/(s-1)

Applying the initial condition and simplifying gives:

(s^2Y+2s+3)-2(sY+2)+Y =(1)/(s-1)

(s^2-2s+1)Y =(1)/(s-1)-2s+1

(s-1)^2Y=(1)/(s-1)-2s+1

Solving for Y:

Y=(1)/((s-1)^3)+(-2s+1)/((s-1)^2) =1/2(2)/((s-1)^3)+(-2s+1)/((s-1)^2)

For the first term, we use: Lap^{:-1:} {(n!)/((s-a)^[n+1])}=e^[at]t^n, with a = 1 and n = 2.

So

Lap^{:-1:}{1/2 (2)/((s-1)^3)} =1/2 e^t t^2

For the second term, we express in partial fractions:

(-2s+1)/((s-1)^2) =(A)/(s-1)+(B)/((s-1)^2)

-2s+1=A(s-1)+B

Comparing coefficients:

-2s=As  gives A = -2.

1=-A+B  gives B = -1.

So (-2s+1)/((s-1)^2) =-(2)/(s-1)-(1)/((s-1)^2)

And

Lap^{:-1:} { - (2)/(s-1) - (1)/((s-1)^2)} =-2e^t - te^t

Putting our inverse Laplace transform expressions together, the solution for y is:

y(t)=1/2 t^2 e^t - 2e^t - te^t

## Application

The current i(t) in an electrical circuit is given by the DE

(d^2i)/(dt^2)+2(di)/(dt)=0,if 0 < t < 10

=1,if 10 < t < 20

=0,if t > 20

and i(0) = 0, i’(0) = 0.

Determine the current as a function of t.

We need to write the RHS of the DE in terms of unit step functions.

(d^2i)/(dt^2)+2(di)/(dt) =u(t-10)-u(t-20)

Now, taking Laplace transform of both sides gives us:

(s^2I-s\ i(0)-i"'"(0))+2(sI-i(0)) =(e^(-10s))/s-(e^(-20s))/s

s^2I+2sI =1/s(e^(-10s)-e^(-20s))

(s^2+2s)I =1/s(e^(-10s)-e^(-20s))

Solving for I gives:

I=1/s((e^(-10s)-e^(-20s))/(s^2+2s)) =1/(s^2(s+2))(e^(-10s)-e^(-20s))

We need to find the Inverse Laplace of this expression. First, we concentrate on the 1/(s^2(s+2)) part and ignore the (e^(-10s)-e^(-20s)) part for now.

Now, we find the partial fractions: 1/(s^2(s+2)) =A/s+B/s^2+C/(s+2)

Multiply both sides by s^2(s+2):

1=As(s+2)+B(s+2)+Cs^2

s = 0 gives 1 = 2B gives B=1/2

s=-2 gives 1 = 4C gives C=1/4

s=1 gives 1 = 3A + 3B + C gives A= -1/4

So 1/(s^2(s+2)) =-1/(4s)+1/(2s^2)+1/(4(s+2))

Now, the inverse Laplace of this expression is:

Lap^{:-1:}{1/(s^2(s+2))}

Lap^{:-1:}{-1/(4s)+1/(2s^2)+1/(4(s+2))}

=-1/4+1/2t+1/4e^(-2t)

So since

I=1/(s^2(s+2))e^(-10s)-1/(s^2(s+2))e^(-20s),

then we have, using the Time-Displacement Theorem (see the Table of Laplace Transforms):

i(t)= [-1/4*u(t-10)+ 1/2(t-10)*u(t-10)+ {:1/4e^(-2(t-10))*u(t-10)]- [-1/4*u(t-20)+ 1/2(t-20)*u(t-20)+ {:1/4e^(-2(t-20))*u(t-20)]

=1/4(2t-21+e^(-2(t-10)))*u(t-10) +1/4(41-2t-e^(-2(t-20)))*u(t-20)

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